Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions (Miscellaneous Exercise) with detailed step-by-step explanations, conceptual clarity, and solved examples as per the latest CBSE and NCERT guidelines. This Miscellaneous Exercise combines concepts from all previous exercises—covering trigonometric identities, general solutions of trigonometric equations, signs of trigonometric functions, values of functions in different quadrants, and relations between degrees and radians. These solutions help students revise and master every concept essential for school exams, JEE, NDA, CUET, and other entrance tests. All questions are solved accurately by expert teachers to strengthen problem-solving skills. Click the print button to download study material and notes in PDF format.
Q.1 : Prove $$2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}=0$$
Solution
LHS
$$
=2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}
$$
Use $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ on the last two terms:
$$
\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}
=2\cos\frac{3\pi/13+5\pi/13}{2}\cos\frac{3\pi/13-5\pi/13}{2}
$$
$$
=2\cos\frac{8\pi}{26}\cos\frac{-2\pi}{26}
=2\cos\frac{4\pi}{13}\cos\left(-\frac{\pi}{13}\right)
=2\cos\frac{4\pi}{13}\cos\frac{\pi}{13}.
$$
So
$$
\text{LHS}=2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+2\cos\frac{4\pi}{13}\cos\frac{\pi}{13}
$$
Factor $2\cos\frac{\pi}{13}$:
$$
\text{LHS}=2\cos\frac{\pi}{13}\big(\cos\frac{9\pi}{13}+\cos\frac{4\pi}{13}\big).
$$
Apply $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ to the bracket:
$$
\cos\frac{9\pi}{13}+\cos\frac{4\pi}{13}
=2\cos\frac{9\pi/13+4\pi/13}{2}\cos\frac{9\pi/13-4\pi/13}{2}
$$
$$
=2\cos\frac{13\pi}{26}\cos\frac{5\pi}{26}
=2\cos\frac{\pi}{2}\cos\frac{5\pi}{26}.
$$
But $\cos\frac{\pi}{2}=0$, therefore the bracket is $0$, so
$$
\text{LHS}=2\cos\frac{\pi}{13}\times 0 \times \cos\frac{5\pi}{26}=0=\text{RHS}.
$$
Hence proved.
Q.2 : Prove $$(\sin 3x + \sin x)\sin x + (\cos 3x – \cos x)\cos x = 0$$
Solution
LHS
$$
=(\sin 3x + \sin x)\sin x + (\cos 3x – \cos x)\cos x
$$
Expand:
$$
=\sin 3x\sin x + \sin^2 x + \cos 3x\cos x – \cos^2 x
$$
Group terms:
$$
=\big(\cos 3x\cos x + \sin 3x\sin x\big) – \big(\cos^2 x – \sin^2 x\big)
$$
Use $\cos A\cos B + \sin A\sin B=\cos(A-B)$ and $\cos^2 A – \sin^2 A=\cos 2A$:
$$
=\cos(3x-x)-\cos 2x
=\cos 2x – \cos 2x = 0.
$$
Hence proved.
Q.3 : Prove $$(\cos x + \cos y)^2 + (\sin x – \sin y)^2 = 4\cos^2\frac{x+y}{2}$$
Solution
LHS
$$
=(\cos x + \cos y)^2 + (\sin x – \sin y)^2
$$
Expand:
$$
=\cos^2 x + \cos^2 y + 2\cos x\cos y
+\sin^2 x + \sin^2 y – 2\sin x\sin y
$$
Group using $\sin^2A+\cos^2A=1$:
$$
=[\sin^2 x+\cos^2 x]+[\sin^2 y+\cos^2 y] + 2(\cos x\cos y – \sin x\sin y)
$$
So
$$
=1+1 + 2\cos(x+y)
=2+2\cos(x+y)
$$
Using $\cos 2A = 2\cos^2 A -1$, write $1+\cos(x+y)=2\cos^2\frac{x+y}{2}$:
$$
2+2\cos(x+y)=2\big[1+\cos(x+y)\big]=2\big[2\cos^2\frac{x+y}{2}\big]
=4\cos^2\frac{x+y}{2}.
$$
Hence proved.
Q.4: Prove $$(\cos x – \cos y)^2 + (\sin x – \sin y)^2 = 4\sin^2\frac{x-y}{2}$$
Solution
LHS
$$
=(\cos x – \cos y)^2 + (\sin x – \sin y)^2
$$
Expand:
$$
=\cos^2 x + \cos^2 y – 2\cos x\cos y
+\sin^2 x + \sin^2 y – 2\sin x\sin y
$$
Group:
$$
=[\sin^2 x+\cos^2 x]+[\sin^2 y+\cos^2 y] – 2(\cos x\cos y + \sin x\sin y)
$$
Use $\sin^2A+\cos^2A=1$ and $\cos x\cos y+\sin x\sin y=\cos(x-y)$:
$$
=1+1 – 2\cos(x-y)
=2 – 2\cos(x-y)
$$
Use $\cos 2A = 1 – 2\sin^2 A$, so $1-\cos(x-y)=2\sin^2\frac{x-y}{2}$:
$$
2 – 2\cos(x-y)=2\big[1-\cos(x-y)\big]
=2\big[2\sin^2\frac{x-y}{2}\big]=4\sin^2\frac{x-y}{2}.
$$
Hence proved.
Q.5 : Prove $$\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos 2x\sin 4x$$
Solution :
LHS
$$
=\sin x + \sin 3x + \sin 5x + \sin 7x
$$
Group as two pairs:
$$
=(\sin x + \sin 5x) + (\sin 3x + \sin 7x)
$$
Use $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$:
$$
\sin x + \sin 5x = 2\sin 3x\cos(-2x)=2\sin 3x\cos 2x,
$$
$$
\sin 3x + \sin 7x = 2\sin 5x\cos(-2x)=2\sin 5x\cos 2x.
$$
So
$$
\text{LHS}=2\sin 3x\cos 2x + 2\sin 5x\cos 2x
=2\cos 2x(\sin 3x + \sin 5x).
$$
Apply $\sin A + \sin B$ to the bracket:
$$
\sin 3x + \sin 5x = 2\sin\frac{3x+5x}{2}\cos\frac{3x-5x}{2}
$$
$$=2\sin 4x\cos(-x)$$
$$=2\sin 4x\cos x.$$
Thus
$$
\text{LHS}=2\cos 2x\cdot (2\sin 4x\cos x)
=4\cos x\cos 2x\sin 4x=\text{RHS}.
$$
Hence proved.
Q.6 : $$\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x$$
Solution :
LHS =
$$\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}$$
Using the identities
$$\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A – B}{2}\right)$$
$$\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A – B}{2}\right)$$
we get:
$$(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)$$
$$= 2 \sin\left(\frac{7x + 5x}{2}\right) \cos\left(\frac{7x – 5x}{2}\right) + 2 \sin\left(\frac{9x + 3x}{2}\right) \cos\left(\frac{9x – 3x}{2}\right)$$
$$= 2 \sin 6x \cos x + 2 \sin 6x \cos 3x$$
$$= 2 \sin 6x (\cos x + \cos 3x)$$
Similarly,
$$(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)$$
$$= 2 \cos\left(\frac{7x + 5x}{2}\right) \cos\left(\frac{7x – 5x}{2}\right) + 2 \cos\left(\frac{9x + 3x}{2}\right) \cos\left(\frac{9x – 3x}{2}\right)$$
$$= 2 \cos 6x \cos x + 2 \cos 6x \cos 3x$$
$$= 2 \cos 6x (\cos x + \cos 3x)$$
Now,
$$
\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)}
= \frac{2 \sin 6x (\cos x + \cos 3x)}{2 \cos 6x (\cos x + \cos 3x)}
$$
$$
= \frac{\sin 6x}{\cos 6x} = \tan 6x = \text{RHS}
$$
Hence, proved.
Q.7 : $$\sin 3x + \sin 2x – \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}$$
Solution :
LHS =
$$\sin 3x + \sin 2x – \sin x$$
Using the identity
$$\sin A – \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A – B}{2}\right),$$
we get:
$$\sin 3x + (\sin 2x – \sin x)$$
$$= \sin 3x + \left[ 2 \cos\left(\frac{2x + x}{2}\right) \sin\left(\frac{2x – x}{2}\right) \right]$$
$$= \sin 3x + 2 \cos\left(\frac{3x}{2}\right) \sin\left(\frac{x}{2}\right)$$
Using $\sin 2A = 2 \sin A \cos A$, we get:
$$\sin 3x = 2 \sin\left(\frac{3x}{2}\right) \cos\left(\frac{3x}{2}\right)$$
Substitute this in LHS:
$$2 \sin\left(\frac{3x}{2}\right) \cos\left(\frac{3x}{2}\right) + 2 \cos\left(\frac{3x}{2}\right) \sin\left(\frac{x}{2}\right)$$
$$= 2 \cos\left(\frac{3x}{2}\right) \left( \sin\left(\frac{3x}{2}\right) + \sin\left(\frac{x}{2}\right) \right)$$
Now using
$$\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A – B}{2}\right),$$
we get:
$$2 \cos\left(\frac{3x}{2}\right) \left[ 2 \sin\left(\frac{3x/2 + x/2}{2}\right) \cos\left(\frac{3x/2 – x/2}{2}\right) \right]$$
$$= 2 \cos\left(\frac{3x}{2}\right) \left[ 2 \sin x \cos\left(\frac{x}{2}\right) \right]$$
$$= 4 \sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)$$
$$= \text{RHS}$$
Hence, proved.
Q.8 : Given: $\tan x = -\dfrac{4}{3}$, $x$ in quadrant II. Find: $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$, $\tan\dfrac{x}{2}$
Solution :
It is given that $x$ is in quadrant II. Therefore,
$$\frac{\pi}{2} < x < \pi$$
Divide by 2:
$$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$
So $\dfrac{x}{2}$ lies in quadrant I.
Hence, $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$ and $\tan\dfrac{x}{2}$ are all positive.
We know that
$$\sec^2 A = 1 + \tan^2 A$$
So,
$$\sec^2 x = 1 + \tan^2 x$$
$$\sec^2 x = 1 + \left(-\frac{4}{3}\right)^2$$
$$\sec^2 x = 1 + \frac{16}{9}$$
$$\sec^2 x = \frac{25}{9}$$
Therefore,
$$\sec x = \pm \frac{5}{3}$$
and
$$\cos x = \pm \frac{3}{5}$$
Since $x$ lies in quadrant II, $\cos x$ is negative.
Hence,
$$\cos x = -\frac{3}{5}$$
We know that
$$\cos x = 2 \cos^2 \frac{x}{2} – 1$$
Substitute the value of $\cos x$:
$$-\frac{3}{5} = 2 \cos^2 \frac{x}{2} – 1$$
Simplify:
$$-\frac{3}{5} + 1 = 2 \cos^2 \frac{x}{2}$$
$$\frac{2}{5} = 2 \cos^2 \frac{x}{2}$$
$$\cos^2 \frac{x}{2} = \frac{1}{5}$$
Since $\frac{x}{2}$ is in quadrant I, $\cos\dfrac{x}{2}$ is positive.
$$\cos \frac{x}{2} = \frac{1}{\sqrt{5}}$$
Now, using $\sin^2 \dfrac{x}{2} + \cos^2 \dfrac{x}{2} = 1$:
$$\sin^2 \frac{x}{2} = 1 – \frac{1}{5}$$
$$\sin^2 \frac{x}{2} = \frac{4}{5}$$
Since $\frac{x}{2}$ lies in quadrant I,
$$\sin \frac{x}{2} = \frac{2}{\sqrt{5}}$$
Now,
$$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$$
$$\tan \frac{x}{2} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}$$
$$\tan \frac{x}{2} = 2$$
Final Answers
$$\sin \frac{x}{2} = \frac{2}{\sqrt{5}}$$
$$\cos \frac{x}{2} = \frac{1}{\sqrt{5}}$$
$$\tan \frac{x}{2} = 2$$
Q.9 : Given: $\cos x = -\dfrac{1}{3}$, $x$ in quadrant III. Find: $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$, $\tan\dfrac{x}{2}$
Solution :
It is given that $x$ lies in quadrant III. Therefore,
$$\pi < x < \frac{3\pi}{2}$$
Divide by 2:
$$\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$$
Hence, $\dfrac{x}{2}$ lies in quadrant II.
So $\sin\dfrac{x}{2}$ is positive, but $\cos\dfrac{x}{2}$ and $\tan\dfrac{x}{2}$ are negative.
We know that
$$\cos 2A = 1 – 2 \sin^2 A$$
So,
$$\cos x = 1 – 2 \sin^2 \frac{x}{2}$$
Substitute $\cos x = -\dfrac{1}{3}$:
$$-\frac{1}{3} = 1 – 2 \sin^2 \frac{x}{2}$$
Simplify:
$$1 + \frac{1}{3} = 2 \sin^2 \frac{x}{2}$$
$$\frac{4}{3} = 2 \sin^2 \frac{x}{2}$$
$$\sin^2 \frac{x}{2} = \frac{2}{3}$$
Since $\frac{x}{2}$ lies in quadrant II, $\sin\dfrac{x}{2} > 0$:
$$\sin \frac{x}{2} = \frac{\sqrt{2}}{\sqrt{3}}$$
Now, use the identity
$$\cos 2A = 2 \cos^2 A – 1$$
So,
$$\cos x = 2 \cos^2 \frac{x}{2} – 1$$
Substitute $\cos x = -\dfrac{1}{3}$:
$$-\frac{1}{3} = 2 \cos^2 \frac{x}{2} – 1$$
Simplify:
$$-\frac{1}{3} + 1 = 2 \cos^2 \frac{x}{2}$$
$$\frac{2}{3} = 2 \cos^2 \frac{x}{2}$$
$$\cos^2 \frac{x}{2} = \frac{1}{3}$$
Since $\frac{x}{2}$ lies in quadrant II, $\cos\dfrac{x}{2}$ is negative:
$$\cos \frac{x}{2} = -\frac{1}{\sqrt{3}}$$
Now,
$$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$$
$$\tan \frac{x}{2} = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{-\frac{1}{\sqrt{3}}}$$
$$\tan \frac{x}{2} = -\sqrt{2}$$
Final Answers
$$\sin \frac{x}{2} = \frac{\sqrt{2}}{\sqrt{3}}$$
$$\cos \frac{x}{2} = -\frac{1}{\sqrt{3}}$$
$$\tan \frac{x}{2} = -\sqrt{2}$$
Q.10 : Given: $\sin x = \dfrac{1}{4}$, $x$ in quadrant II. Find: $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$, $\tan\dfrac{x}{2}$
Solution :
It is given that $x$ lies in quadrant II. Therefore,
$$\frac{\pi}{2} < x < \pi$$
Divide by 2:
$$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$
Hence, $\dfrac{x}{2}$ lies in quadrant I.
So $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$, and $\tan\dfrac{x}{2}$ are all positive.
We know that
$$\cos^2 A = 1 – \sin^2 A$$
So,
$$\cos^2 x = 1 – \sin^2 x$$
$$\cos^2 x = 1 – \left(\frac{1}{4}\right)^2$$
$$\cos^2 x = 1 – \frac{1}{16}$$
$$\cos^2 x = \frac{15}{16}$$
Hence,
$$\cos x = \pm \frac{\sqrt{15}}{4}$$
Since $x$ is in quadrant II, $\cos x < 0$:
$$\cos x = -\frac{\sqrt{15}}{4}$$
Using the half-angle identity:
$$\cos x = 2 \cos^2 \frac{x}{2} – 1$$
Substitute $\cos x = -\frac{\sqrt{15}}{4}$:
$$-\frac{\sqrt{15}}{4} = 2 \cos^2 \frac{x}{2} – 1$$
Simplify:
$$2 \cos^2 \frac{x}{2} = 1 + \frac{\sqrt{15}}{4}$$
$$2 \cos^2 \frac{x}{2} = \frac{4 + \sqrt{15}}{4}$$
$$\cos^2 \frac{x}{2} = \frac{4 + \sqrt{15}}{8}$$
Since $\frac{x}{2}$ is in quadrant I, $\cos\dfrac{x}{2} > 0$:
$$\cos \frac{x}{2} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}$$
Now,
$$\sin^2 \frac{x}{2} = 1 – \cos^2 \frac{x}{2}$$
$$\sin^2 \frac{x}{2} = 1 – \frac{4 + \sqrt{15}}{8}$$
$$\sin^2 \frac{x}{2} = \frac{4 – \sqrt{15}}{8}$$
Hence,
$$\sin \frac{x}{2} = \frac{\sqrt{4 – \sqrt{15}}}{2\sqrt{2}}$$
Finally,
$$\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$$
$$\tan \frac{x}{2} = \frac{\frac{\sqrt{4 – \sqrt{15}}}{2\sqrt{2}}}{\frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}}$$
$$\tan \frac{x}{2} = \frac{\sqrt{4 – \sqrt{15}}}{\sqrt{4 + \sqrt{15}}}$$
Final Answers
$$\sin \frac{x}{2} = \frac{\sqrt{4 – \sqrt{15}}}{2\sqrt{2}}$$
$$\cos \frac{x}{2} = \frac{\sqrt{4 + \sqrt{15}}}{2\sqrt{2}}$$
$$\tan \frac{x}{2} = \frac{\sqrt{4 – \sqrt{15}}}{\sqrt{4 + \sqrt{15}}}$$
