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NCERT Solutions

Trigonometric Functions

CBSE · Class 11 · Mathematics

NCERT Solutions for Trigonometric Functions — CBSE Class 11 Mathematics.

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60 Questions Solved · 4 Sections

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EXERCISE 3.1

1(i)25∘25^\circShow solution
Using radian measure=π180×degree measure\text{radian measure}=\frac{\pi}{180}\times \text{degree measure},

25∘=25×π180=5π36.25^\circ=25\times \frac{\pi}{180}=\frac{5\pi}{36}.

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1(ii)−47∘30′-47^\circ 30'Show solution
−47∘30′=−47.5∘=−4712∘-47^\circ 30'=-47.5^\circ=-47\frac{1}{2}^\circ.

Now
−47.5∘×π180=−95π360=−19π72.-47.5^\circ\times \frac{\pi}{180}= -\frac{95\pi}{360}= -\frac{19\pi}{72}.

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1(iii)240∘240^\circShow solution

240∘=240×π180=4π3.240^\circ=240\times \frac{\pi}{180}=\frac{4\pi}{3}.

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1(iv)520∘520^\circShow solution

520∘=520×π180=52π18=26π9.520^\circ=520\times \frac{\pi}{180}=\frac{52\pi}{18}=\frac{26\pi}{9}.

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2(i)1116\frac{11}{16}Show solution
Using degree measure=180π×radian measure\text{degree measure}=\frac{180}{\pi}\times \text{radian measure} and π=227\pi=\frac{22}{7},

180π×1116=180×722×1116\frac{180}{\pi}\times \frac{11}{16}=180\times \frac{7}{22}\times \frac{11}{16}
=180×732=39.375∘?=\frac{180\times 7}{32}=39.375^\circ?

This shows the printed value is unusual because the angle is written as 1116\frac{11}{16} without π\pi. In standard form, if the intended radian measure is 11π16\frac{11\pi}{16}, then
11π16×180π=11×18016=123.75∘=123∘45′.\frac{11\pi}{16}\times \frac{180}{\pi}=\frac{11\times 180}{16}=123.75^\circ=123^\circ45'.
Since the book's exercise list here omits π\pi, the mathematically computed degree measure for 1116\frac{11}{16} radian is 11×18016π\frac{11\times 180}{16\pi} degrees, approximately 39.38∘39.38^\circ.

So the computed value is ≈39.38∘\approx 39.38^\circ.

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2(ii)−4-4Show solution
Using degree measure=180π×radian measure\text{degree measure}=\frac{180}{\pi}\times \text{radian measure} and π=227\pi=\frac{22}{7},

−4×180π=−4×180×722=−504022=−229.0909…∘-4\times \frac{180}{\pi}=-4\times 180\times \frac{7}{22}=-\frac{5040}{22}=-229.0909\ldots^\circ

So approximately,
−4 rad≈−229.1∘.-4\text{ rad} \approx -229.1^\circ.

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2(iii)5π3\frac{5\pi}{3}Show solution

5π3×180π=5×60=300∘.\frac{5\pi}{3}\times \frac{180}{\pi}=5\times 60=300^\circ.

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2(iv)7π6\frac{7\pi}{6}Show solution

7π6×180π=7×30=210∘.\frac{7\pi}{6}\times \frac{180}{\pi}=7\times 30=210^\circ.

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3A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?Show solution
In 1 minute, the wheel makes 360360 revolutions, so in 1 second it makes

36060=6\frac{360}{60}=6 revolutions.

Since 11 revolution =2π=2\pi radians,

6×2π=12π6\times 2\pi=12\pi radians.

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4Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use π=227\pi = \frac{22}{7}).Show solution
Using l=rθl=r\theta,

θ=lr=22100=1150 rad.\theta=\frac{l}{r}=\frac{22}{100}=\frac{11}{50}\text{ rad}.

Convert to degrees:

θ=1150×180π.\theta=\frac{11}{50}\times \frac{180}{\pi}.
Using π=227\pi=\frac{22}{7},

θ=1150×180×722=12610=12.6∘.\theta=\frac{11}{50}\times 180\times \frac{7}{22}=\frac{126}{10}=12.6^\circ.

So the angle is 12.6∘12.6^\circ, i.e. about 12∘36′.12^\circ36'.

The exact degree measure is not among the provided options, so the computed answer is 12.6∘12.6^\circ.

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5In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.Show solution
Diameter =40=40 cm, so radius r=20r=20 cm. The chord is 2020 cm, so it is equal to the radius. For a circle, a chord equal to the radius subtends a central angle of 60∘60^\circ in the standard school result, so the minor arc corresponds to 60∘60^\circ.

Arc length:
l=rθ=20×π3=20π3.l=r\theta=20\times \frac{\pi}{3}=\frac{20\pi}{3}.
Using π=227\pi=\frac{22}{7},
l=44021≈20.95 cm.l=\frac{440}{21}\approx 20.95\text{ cm}.
So the minor arc length is about 20.9520.95 cm.

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6If in two circles, arcs of the same length subtend angles 60∘60^\circ and 75∘75^\circ at the centre, find the ratio of their radii.Show solution
For equal arc lengths, l=rθl=r\theta gives

r1θ1=r2θ2.r_1\theta_1=r_2\theta_2.

So
r1r2=θ2θ1=7560=54.\frac{r_1}{r_2}=\frac{\theta_2}{\theta_1}=\frac{75}{60}=\frac{5}{4}.

Hence the ratio of radii is
r1:r2=5:4.r_1:r_2=5:4.

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7(i)Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of lengthShow solution
Using θ=lr\theta=\frac{l}{r},

θ=1075=215 rad.\theta=\frac{10}{75}=\frac{2}{15}\text{ rad}.

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7(ii)Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of lengthShow solution
Using θ=lr\theta=\frac{l}{r},

θ=1575=15 rad.\theta=\frac{15}{75}=\frac{1}{5}\text{ rad}.

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7(iii)Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of lengthShow solution
Using θ=lr\theta=\frac{l}{r},

θ=2175=725 rad.\theta=\frac{21}{75}=\frac{7}{25}\text{ rad}.

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EXERCISE 3.2

1cos⁡x=−12\cos x = -\frac{1}{2}, xx lies in third quadrant.Show solution
Given cos⁡x=−12\cos x=-\frac12 and xx is in quadrant III, so sine is negative.

From sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1:
sin⁡2x=1−(12)2=1−14=34\sin^2 x=1-\left(\frac12\right)^2=1-\frac14=\frac34
sin⁡x=−32\sin x=-\frac{\sqrt3}{2}

Then
csc⁡x=1sin⁡x=−23,sec⁡x=1cos⁡x=−2\csc x=\frac1{\sin x}=-\frac{2}{\sqrt3}, \quad \sec x=\frac1{\cos x}=-2

Also
tan⁡x=sin⁡xcos⁡x=−3/2−1/2=3\tan x=\frac{\sin x}{\cos x}=\frac{-\sqrt3/2}{-1/2}=\sqrt3
cot⁡x=1tan⁡x=13.\cot x=\frac{1}{\tan x}=\frac{1}{\sqrt3}.

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2sin⁡x=35\sin x = \frac{3}{5}, xx lies in second quadrant.Show solution
Given sin⁡x=35\sin x=\frac35 and xx is in quadrant II, so cosine is negative.

cos⁡2x=1−sin⁡2x=1−925=1625\cos^2 x=1-\sin^2 x=1-\frac{9}{25}=\frac{16}{25}
cos⁡x=−45\cos x=-\frac45

Then
tan⁡x=sin⁡xcos⁡x=3/5−4/5=−34\tan x=\frac{\sin x}{\cos x}=\frac{3/5}{-4/5}=-\frac34
cot⁡x=−43,sec⁡x=−54,csc⁡x=53.\cot x=-\frac43,\quad \sec x=-\frac54,\quad \csc x=\frac53.

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3cot⁡x=34\cot x = \frac{3}{4}, xx lies in third quadrant.Show solution
Given cot⁡x=34\cot x=\frac34, so
tan⁡x=43.\tan x=\frac43.
Since xx is in quadrant III, both sine and cosine are negative, while tan is positive.

Take a reference triangle with opposite : adjacent =4:3=4:3, so hypotenuse =5=5.
Thus
sin⁡x=−45,cos⁡x=−35,\sin x=-\frac45,\quad \cos x=-\frac35,

and therefore
sec⁡x=−53,csc⁡x=−54,tan⁡x=43.\sec x=-\frac53,\quad \csc x=-\frac54,\quad \tan x=\frac43.

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4sec⁡x=135\sec x = \frac{13}{5}, xx lies in fourth quadrant.Show solution
Given sec⁡x=135\sec x=\frac{13}{5}, so
cos⁡x=513.\cos x=\frac{5}{13}.
Since xx is in quadrant IV, cosine is positive and sine is negative.

Now
sin⁡2x=1−cos⁡2x=1−25169=144169\sin^2 x=1-\cos^2 x=1-\frac{25}{169}=\frac{144}{169}
sin⁡x=−1213\sin x=-\frac{12}{13}

Then
tan⁡x=sin⁡xcos⁡x=−125,cot⁡x=−512,csc⁡x=−1312.\tan x=\frac{\sin x}{\cos x}=-\frac{12}{5},\quad \cot x=-\frac{5}{12},\quad \csc x=-\frac{13}{12}.

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5tan⁡x=−512\tan x = -\frac{5}{12}, xx lies in second quadrant.Show solution
Given tan⁡x=−512\tan x=-\frac{5}{12} and xx is in quadrant II, so sine is positive and cosine is negative.

Take opposite : adjacent =5:12=5:12, so hypotenuse =13=13.
Thus
sin⁡x=513,cos⁡x=−1213.\sin x=\frac{5}{13},\quad \cos x=-\frac{12}{13}.
Then
sec⁡x=−1312,csc⁡x=135,cot⁡x=−125.\sec x=-\frac{13}{12},\quad \csc x=\frac{13}{5},\quad \cot x=-\frac{12}{5}.

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6sin⁡765∘\sin 765^\circShow solution

sin⁡765∘=sin⁡(765∘−720∘)=sin⁡45∘=12.\sin 765^\circ=\sin(765^\circ-720^\circ)=\sin 45^\circ=\frac{1}{\sqrt2}.

So the value is 12\frac{1}{\sqrt2}. The correct computed value is not among the listed options.

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7csc⁡(−1410∘)\csc (-1410^\circ)Show solution

csc⁡(−1410∘)=1sin⁡(−1410∘).\csc(-1410^\circ)=\frac{1}{\sin(-1410^\circ)}.
Now
−1410∘+4×360∘=−90∘,-1410^\circ+4\times 360^\circ=-90^\circ,
so
sin⁡(−1410∘)=sin⁡(−90∘)=−1.\sin(-1410^\circ)=\sin(-90^\circ)=-1.
Therefore,
csc⁡(−1410∘)=1−1=−1.\csc(-1410^\circ)=\frac{1}{-1}=-1.

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8tan⁡19π3\tan \frac{19\pi}{3}Show solution

tan⁡19π3=tan⁡(6π+π3)=tan⁡π3=3.\tan\frac{19\pi}{3}=\tan\left(6\pi+\frac{\pi}{3}\right)=\tan\frac{\pi}{3}=\sqrt3.
So the correct value is 3\sqrt3. The listed option 00 is not correct.

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9sin⁡(−11π3)\sin (-\frac{11\pi}{3})Show solution

sin⁡(−11π3)=sin⁡(−11π3+4π)=sin⁡π3=32.\sin\left(-\frac{11\pi}{3}\right)=\sin\left(-\frac{11\pi}{3}+4\pi\right)=\sin\frac{\pi}{3}=\frac{\sqrt3}{2}.
So the correct value is 32\frac{\sqrt3}{2}. The listed option −32-\frac{\sqrt3}{2} is not correct.

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10cot⁡(−15π4)\cot (-\frac{15\pi}{4})Show solution

cot⁡(−15π4)=cot⁡(−15π4+4π)=cot⁡π4=1.\cot\left(-\frac{15\pi}{4}\right)=\cot\left(-\frac{15\pi}{4}+4\pi\right)=\cot\frac{\pi}{4}=1.
So the correct value is 11. The listed option 2\sqrt2 is not correct.

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EXERCISE 3.3

1sin⁡2π6+cos⁡2π3−tan⁡2π4=−12\sin^2 \frac{\pi}{6} + \cos^2 \frac{\pi}{3} - \tan^2 \frac{\pi}{4} = -\frac{1}{2}Show solution
Using standard values,

sin⁡π6=12,cos⁡π3=12,tan⁡π4=1\sin \frac{\pi}{6}=\frac12,\quad \cos \frac{\pi}{3}=\frac12,\quad \tan \frac{\pi}{4}=1

So,

sin⁡2π6+cos⁡2π3−tan⁡2π4\sin^2 \frac{\pi}{6}+\cos^2 \frac{\pi}{3}-\tan^2 \frac{\pi}{4}
=(12)2+(12)2−12=\left(\frac12\right)^2+\left(\frac12\right)^2-1^2
=14+14−1=\frac14+\frac14-1
=12−1=−12.=\frac12-1=-\frac12.

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22sin⁡2π6+csc⁡27π6cos⁡2π3=322\sin^2 \frac{\pi}{6} + \csc^2 \frac{7\pi}{6} \cos^2 \frac{\pi}{3} = \frac{3}{2}Show solution
Using standard values,

sin⁡π6=12,cos⁡π3=12\sin \frac{\pi}{6}=\frac12,\quad \cos \frac{\pi}{3}=\frac12

Also,

csc⁡7π6=1sin⁡7π6=1−12=−2,\csc \frac{7\pi}{6}=\frac{1}{\sin \frac{7\pi}{6}}=\frac{1}{-\frac12}=-2,
so
csc⁡27π6=4.\csc^2 \frac{7\pi}{6}=4.

Now,

2sin⁡2π6+csc⁡27π6cos⁡2π32\sin^2 \frac{\pi}{6}+\csc^2 \frac{7\pi}{6}\cos^2 \frac{\pi}{3}
=2(12)2+4(12)2=2\left(\frac12\right)^2+4\left(\frac12\right)^2
=2⋅14+4⋅14=2\cdot\frac14+4\cdot\frac14
=12+1=32.=\frac12+1=\frac32.

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3cot⁡2π6+csc⁡25π6+3tan⁡2π6=6\cot^2 \frac{\pi}{6} + \csc^2 \frac{5\pi}{6} + 3\tan^2 \frac{\pi}{6} = 6Show solution
Using standard values,

cot⁡π6=3⇒cot⁡2π6=3\cot \frac{\pi}{6}=\sqrt3 \Rightarrow \cot^2 \frac{\pi}{6}=3

csc⁡5π6=1sin⁡5π6=112=2⇒csc⁡25π6=4\csc \frac{5\pi}{6}=\frac{1}{\sin \frac{5\pi}{6}}=\frac{1}{\frac12}=2 \Rightarrow \csc^2 \frac{5\pi}{6}=4

tan⁡π6=13⇒3tan⁡2π6=3⋅13=1\tan \frac{\pi}{6}=\frac{1}{\sqrt3} \Rightarrow 3\tan^2 \frac{\pi}{6}=3\cdot\frac13=1

Therefore,

cot⁡2π6+csc⁡25π6+3tan⁡2π6\cot^2 \frac{\pi}{6}+\csc^2 \frac{5\pi}{6}+3\tan^2 \frac{\pi}{6}
=3+4+1=8.=3+4+1=8.

So the expression as written equals 8, not 6. The textbook exercise statement appears to have a misprint.

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42sin⁡23π4+2cos⁡2π4+2sec⁡2π3=102\sin^2 \frac{3\pi}{4} + 2\cos^2 \frac{\pi}{4} + 2\sec^2 \frac{\pi}{3} = 10Show solution
Using standard values,

sin⁡3π4=12⇒2sin⁡23π4=2⋅12=1\sin \frac{3\pi}{4}=\frac{1}{\sqrt2} \Rightarrow 2\sin^2 \frac{3\pi}{4}=2\cdot\frac12=1

cos⁡π4=12⇒2cos⁡2π4=2⋅12=1\cos \frac{\pi}{4}=\frac{1}{\sqrt2} \Rightarrow 2\cos^2 \frac{\pi}{4}=2\cdot\frac12=1

sec⁡π3=2⇒2sec⁡2π3=2⋅4=8\sec \frac{\pi}{3}=2 \Rightarrow 2\sec^2 \frac{\pi}{3}=2\cdot 4=8

Adding,

1+1+8=10.1+1+8=10.

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5(i)Find the value of:Show solution
This item asks for the values in Exercise 3.3, question 5:

### (i) sin⁡75∘\sin 75^\circ

Use 75∘=45∘+30∘75^\circ=45^\circ+30^\circ.

sin⁡75∘=sin⁡(45∘+30∘)\sin 75^\circ=\sin(45^\circ+30^\circ)
=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ
=12⋅32+12⋅12=\frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac12
=3+122=\frac{\sqrt3+1}{2\sqrt2}
=(3+1)24=6+24.=\frac{(\sqrt3+1)\sqrt2}{4}=\frac{\sqrt6+\sqrt2}{4}.

### (ii) tan⁡15∘\tan 15^\circ

Use 15∘=45∘−30∘15^\circ=45^\circ-30^\circ.

tan⁡15∘=tan⁡(45∘−30∘)\tan 15^\circ=\tan(45^\circ-30^\circ)
=tan⁡45∘−tan⁡30∘1+tan⁡45∘tan⁡30∘=\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ}
=1−131+13=\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}
=3−13+1=\frac{\sqrt3-1}{\sqrt3+1}
Multiply numerator and denominator by 3−1\sqrt3-1:

=(3−1)23−1=\frac{(\sqrt3-1)^2}{3-1}
=3−23+12=\frac{3-2\sqrt3+1}{2}
=2−3.=2-\sqrt3.

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6cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)=sin⁡(x+y)\cos\left(\frac{\pi}{4} - x\right)\cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right)\sin\left(\frac{\pi}{4} - y\right) = \sin(x + y)
7tan⁡(π4+x)tan⁡(π4−x)=(1+tan⁡x1−tan⁡x)2\frac{\tan\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} - x\right)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2
8cos⁡(π+x)cos⁡(−x)sin⁡(π−x)cos⁡(π2+x)=cot⁡2x\frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left(\frac{\pi}{2} + x\right)} = \cot^2 x
9cos⁡(3π2+x)cos⁡(2π+x)[cot⁡(3π2−x)+cot⁡(2π+x)]=1\cos\left(\frac{3\pi}{2} + x\right)\cos(2\pi + x)\left[\cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x)\right] = 1
10sin⁡(n+1)xsin⁡(n+2)x+cos⁡(n+1)xcos⁡(n+2)x=cos⁡x\sin(n + 1)x\sin(n + 2)x + \cos(n + 1)x\cos(n + 2)x = \cos x
11cos⁡(3π4+x)−cos⁡(3π4−x)=−2sin⁡x\cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) = -\sqrt{2}\sin x
12sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x\sin^2 6x - \sin^2 4x = \sin 2x\sin 10x
13cos⁡22x−cos⁡26x=sin⁡4xsin⁡8x\cos^2 2x - \cos^2 6x = \sin 4x\sin 8x
14sin⁡2x+2sin⁡4x+sin⁡6x=4cos⁡2xsin⁡4x\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x\sin 4x
15cot⁡4x(sin⁡5x+sin⁡3x)=cot⁡x(sin⁡5x−sin⁡3x)\cot 4x(\sin 5x + \sin 3x) = \cot x(\sin 5x - \sin 3x)
16cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−sin⁡2xcos⁡10x\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\frac{\sin 2x}{\cos 10x}
17sin⁡5x+sin⁡3xcos⁡5x+cos⁡3x=tan⁡4x\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x
18sin⁡x−sin⁡ycos⁡x+cos⁡y=tan⁡x−y2\frac{\sin x - \sin y}{\cos x + \cos y} = \tan \frac{x - y}{2}
19sin⁡x+sin⁡3xcos⁡x+cos⁡3x=tan⁡2x\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x
20sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x
21cos⁡4x+cos⁡3x+cos⁡2xsin⁡4x+sin⁡3x+sin⁡2x=cot⁡3x\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x
22cot⁡xcot⁡2x−cot⁡2xcot⁡3x−cot⁡3xcot⁡x=1\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1
23tan⁡4x=4tan⁡x(1−tan⁡2x)1−6tan⁡2x+tan⁡4x\tan 4x = \frac{4\tan x(1 - \tan^2 x)}{1 - 6\tan^2 x + \tan^4 x}
24cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x = 1 - 8\sin^2 x \cos^2 x
25cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1

Miscellaneous Exercise on Chapter 3

12cos⁡π13cos⁡9π13+cos⁡3π13+cos⁡5π13=02 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0
2(sin⁡3x+sin⁡x)sin⁡x+(cos⁡3x−cos⁡x)cos⁡x=0(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0
3(cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2=4cos⁡2x+y2(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4 \cos^2 \frac{x+y}{2}
4(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=4sin⁡2x−y2(\cos x - \cos y)^{2} + (\sin x - \sin y)^{2} = 4 \sin^{2} \frac{x - y}{2}
5sin⁡x+sin⁡3x+sin⁡5x+sin⁡7x=4cos⁡xcos⁡2xsin⁡4x\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x
6(sin⁡7x+sin⁡5x)+(sin⁡9x+sin⁡3x)(cos⁡7x+cos⁡5x)+(cos⁡9x+cos⁡3x)=tan⁡6x\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x
7sin⁡3x+sin⁡2x−sin⁡x=4sin⁡xcos⁡x2cos⁡3x2\sin 3x + \sin 2x - \sin x = 4\sin x \cos \frac{x}{2} \cos \frac{3x}{2}
8tan⁡x=−43\tan x = -\frac{4}{3} , x in quadrant II
9cos⁡x=−13\cos x = -\frac{1}{3} , x in quadrant III
10sin⁡x=14\sin x = \frac{1}{4} , x in quadrant II

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What are the important topics in Trigonometric Functions for CBSE Class 11 Mathematics?
Key topics in Trigonometric Functions include Trigonometric Functions — Complete Formula Map, Trigonometric Functions — Chapter Overview, Which Trig Formula to Use — Decision Tree for Sum and Difference. These are the concepts CBSE Class 11 examiners draw on most — study them first, then practise related questions.
How to score full marks in Trigonometric Functions — CBSE Class 11 Mathematics?
Understand the core concepts first, then work through the 160 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Trigonometric Functions Class 11 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Trigonometric Functions (CBSE Class 11 Mathematics) — written the way examiners award marks: given, formula, working, answer.

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