NCERT Exemplar Solutions Class 11 Maths Trigonometric Functions — Long Answer Type Q20–29

NCERT Exemplar Problems, Class 11 Mathematics — Trigonometric Functions, Long Answer Type questions 20 to 29. All 10 questions solved.

Ten longer questions. Questions 20 to 24 are proofs that need a clever substitution (writing $\alpha-\beta$ as $(\theta+\alpha)-(\theta+\beta)$, componendo and dividendo, the $t=\tan\theta$ substitution, eliminating $\phi$). Questions 22, 25 and 26 are exact-value computations. Questions 27 to 29 solve trigonometric equations.

Two answers in the printed key are wrong, Questions 28 and 29. Each is explained where it occurs.

Key insight. When the question gives you a relation between sines or cosines of shifted angles, do not expand blindly: look for the angle that cancels the unknown ($\theta$ in Question 20, $\phi$ in Question 24), and for equations, reduce to one function before solving.

Question 20

If $\sin(\theta+\alpha)=a$ and $\sin(\theta+\beta)=b$, then prove that $\cos2(\alpha-\beta)-4ab\cos(\alpha-\beta)=1-2a^2-2b^2$.

Solution. Follow the hint: $\alpha-\beta=(\theta+\alpha)-(\theta+\beta)$, which removes $\theta$. Let $C=\cos(\alpha-\beta)$. Then $$C=\cos(\theta+\alpha)\cos(\theta+\beta)+\sin(\theta+\alpha)\sin(\theta+\beta)=\cos(\theta+\alpha)\cos(\theta+\beta)+ab.$$ Since $\cos2(\alpha-\beta)=2C^2-1$, the left side is $2C^2-1-4abC$, and the claim is $2C^2-1-4abC=1-2a^2-2b^2$, that is $$C^2-2abC+a^2+b^2-1=0\iff(C-ab)^2=a^2b^2-a^2-b^2+1=(1-a^2)(1-b^2).$$ But $C-ab=\cos(\theta+\alpha)\cos(\theta+\beta)$, so $(C-ab)^2=\cos^2(\theta+\alpha)\cos^2(\theta+\beta)=(1-a^2)(1-b^2)$, as required.

Question 21

If $\cos(\theta+\phi)=m\cos(\theta-\phi)$, then prove that $\tan\theta=\dfrac{1-m}{1+m}\cot\phi$.

Solution. Expand both sides: $\cos\theta\cos\phi-\sin\theta\sin\phi=m\cos\theta\cos\phi+m\sin\theta\sin\phi$. Collect the $\cos\theta\cos\phi$ terms on one side and the $\sin\theta\sin\phi$ terms on the other: $$(1-m)\cos\theta\cos\phi=(1+m)\sin\theta\sin\phi.$$ Divide by $(1+m)\cos\theta\sin\phi$: $\tan\theta=\dfrac{1-m}{1+m}\cdot\dfrac{\cos\phi}{\sin\phi}=\dfrac{1-m}{1+m}\cot\phi$. (The hint’s route through componendo and dividendo on $\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}=\frac m1$ gives the same result.)

Question 22

Find the value of the expression $3\left[\sin^4\left(\dfrac{3\pi}2-\alpha\right)+\sin^4(3\pi+\alpha)\right]-2\left\{\sin^6\left(\dfrac\pi2+\alpha\right)+\sin^6(5\pi-\alpha)\right\}$.

Solution. Reduce each angle to a standard one. $\sin\left(\frac{3\pi}2-\alpha\right)=-\cos\alpha$, $\sin(3\pi+\alpha)=-\sin\alpha$, $\sin\left(\frac\pi2+\alpha\right)=\cos\alpha$, $\sin(5\pi-\alpha)=\sin\alpha$. The even powers remove the minus signs, so with $s=\sin\alpha$, $c=\cos\alpha$ the expression is $$3(c^4+s^4)-2(c^6+s^6).$$ Use $c^4+s^4=(c^2+s^2)^2-2s^2c^2=1-2s^2c^2$ and $c^6+s^6=(c^2+s^2)^3-3s^2c^2(c^2+s^2)=1-3s^2c^2$: $$3(1-2s^2c^2)-2(1-3s^2c^2)=3-6s^2c^2-2+6s^2c^2=1.$$

Question 23

If $a\cos2\theta+b\sin2\theta=c$ has $\alpha$ and $\beta$ as its roots, then prove that $\tan\alpha+\tan\beta=\dfrac{2b}{a+c}$.

Solution. Put $t=\tan\theta$ and use the hint: $\cos2\theta=\dfrac{1-t^2}{1+t^2}$ and $\sin2\theta=\dfrac{2t}{1+t^2}$. The equation becomes $$a(1-t^2)+2bt=c(1+t^2)\iff(a+c)t^2-2bt+(c-a)=0.$$ The roots $\alpha$, $\beta$ of the original equation (two values of $\theta$ not differing by a multiple of $\pi$) give two values $\tan\alpha$, $\tan\beta$ of $t$, which are the two roots of this quadratic. The sum of the roots is $-\dfrac{-2b}{a+c}$: $$\tan\alpha+\tan\beta=\frac{2b}{a+c}.$$

Question 24

If $x=\sec\phi-\tan\phi$ and $y=\mathrm{cosec}\,\phi+\cot\phi$, then show that $xy+x-y+1=0$.

Solution. Follow the hint: show $x-y=-(xy+1)$. Write everything over $\sin\phi\cos\phi$, with $s=\sin\phi$, $c=\cos\phi$: $x=\dfrac{1-s}c$ and $y=\dfrac{1+c}s$. Then $$xy+1=\frac{(1-s)(1+c)+sc}{sc}=\frac{1+c-s-sc+sc}{sc}=\frac{1+c-s}{sc},$$ $$x-y=\frac{s(1-s)-c(1+c)}{sc}=\frac{s-c-(s^2+c^2)}{sc}=\frac{s-c-1}{sc}=-\frac{1+c-s}{sc}.$$ So $x-y=-(xy+1)$, which rearranges to $xy+x-y+1=0$.

Question 25

If $\theta$ lies in the first quadrant and $\cos\theta=\dfrac8{17}$, then find the value of $\cos(30^\circ+\theta)+\cos(45^\circ-\theta)+\cos(120^\circ-\theta)$.

Solution. In the first quadrant $\sin\theta=\sqrt{1-\frac{64}{289}}=\frac{15}{17}$. Expand each term with $\cos(A\pm B)$: $$\cos(30^\circ+\theta)=\frac{\sqrt3}2\cdot\frac8{17}-\frac12\cdot\frac{15}{17}=\frac{8\sqrt3-15}{34},$$ $$\cos(45^\circ-\theta)=\frac1{\sqrt2}\left(\frac8{17}+\frac{15}{17}\right)=\frac{23}{17\sqrt2}=\frac{23\sqrt2}{34},$$ $$\cos(120^\circ-\theta)=-\frac12\cdot\frac8{17}+\frac{\sqrt3}2\cdot\frac{15}{17}=\frac{-8+15\sqrt3}{34}.$$ Adding, the constants $-15$ and $-8$ combine to $-23$ and the surds combine to $23\sqrt3+23\sqrt2$: $$\frac{23(\sqrt3+\sqrt2-1)}{34}=\frac{23}{17}\left(\frac{\sqrt3-1}2+\frac1{\sqrt2}\right),$$ which is the form printed in the key.

Question 26

Find the value of the expression $\cos^4\dfrac\pi8+\cos^4\dfrac{3\pi}8+\cos^4\dfrac{5\pi}8+\cos^4\dfrac{7\pi}8$.

Solution. Pair the terms: $\cos\frac{7\pi}8=-\cos\frac\pi8$ and $\cos\frac{5\pi}8=-\cos\frac{3\pi}8$, and even powers remove the signs. So the sum is $2\left(\cos^4\frac\pi8+\cos^4\frac{3\pi}8\right)$. Next, $\cos\frac{3\pi}8=\sin\frac\pi8$, so with $c=\cos\frac\pi8$, $s=\sin\frac\pi8$: $$2(c^4+s^4)=2\left[(c^2+s^2)^2-2c^2s^2\right]=2-4c^2s^2=2-\sin^2\frac\pi4=2-\frac12=\frac32.$$

Question 27

Find the general solution of the equation $5\cos^2\theta+7\sin^2\theta-6=0$.

Syllabus note: general solutions of trigonometric equations are no longer in the current NCERT Class 11 syllabus.

Solution. Write $5\cos^2\theta=5-5\sin^2\theta$: $5+2\sin^2\theta-6=0$, so $\sin^2\theta=\frac12$. This is $\sin^2\theta=\sin^2\frac\pi4$, and $\sin^2\theta=\sin^2\alpha$ has general solution $\theta=n\pi\pm\alpha$.

Question 28

Find the general solution of the equation $\sin x-3\sin2x+\sin3x=\cos x-3\cos2x+\cos3x$.

Syllabus note: general solutions of trigonometric equations are no longer in the current NCERT Class 11 syllabus.

Solution. Group the outer terms on each side with sum-to-product: $\sin3x+\sin x=2\sin2x\cos x$ and $\cos3x+\cos x=2\cos2x\cos x$. The equation becomes $$2\sin2x\cos x-3\sin2x=2\cos2x\cos x-3\cos2x\iff\sin2x\,(2\cos x-3)=\cos2x\,(2\cos x-3).$$ The factor $2\cos x-3$ is never zero (it lies between $-5$ and $-1$), so divide it out: $\sin2x=\cos2x$, i.e. $\tan2x=1=\tan\frac\pi4$. Hence $2x=n\pi+\frac\pi4$ and $$x=\frac{n\pi}2+\frac\pi8.$$

The printed key gives $\frac{n\pi}2\pm\frac\pi8$; the minus branch is wrong. Take $x=-\frac\pi8$ ($n=0$ with the minus sign): then $2x=-\frac\pi4$, so $\tan2x=-1$, not $1$, and the equation fails. Every solution is of the form $\frac{n\pi}2+\frac\pi8$ only.

Question 29

Find the general solution of the equation $(\sqrt3-1)\cos\theta+(\sqrt3+1)\sin\theta=2$.

Syllabus note: general solutions of trigonometric equations are no longer in the current NCERT Class 11 syllabus.

Solution. Follow the hint: put $\sqrt3-1=r\sin\alpha$ and $\sqrt3+1=r\cos\alpha$. Then $r^2=(\sqrt3-1)^2+(\sqrt3+1)^2=8$, so $r=2\sqrt2$, and $\tan\alpha=\dfrac{\sqrt3-1}{\sqrt3+1}=\tan\left(\frac\pi4-\frac\pi6\right)$, so $\alpha=\frac\pi{12}$. The left side is $$r(\sin\alpha\cos\theta+\cos\alpha\sin\theta)=2\sqrt2\sin(\theta+\alpha)=2\quad\Rightarrow\quad\sin\left(\theta+\frac\pi{12}\right)=\frac1{\sqrt2}=\sin\frac\pi4.$$ So $\theta+\frac\pi{12}=n\pi+(-1)^n\frac\pi4$ and $$\theta=n\pi+(-1)^n\frac\pi4-\frac\pi{12}.$$ Split by parity: for $n=2k$, $\theta=2k\pi+\frac\pi6$; for $n=2k+1$, $\theta=2k\pi+\pi-\frac\pi4-\frac\pi{12}=2k\pi+\frac{2\pi}3$. Check: $\theta=\frac\pi6$ gives $(\sqrt3-1)\frac{\sqrt3}2+(\sqrt3+1)\frac12=\frac{3-\sqrt3+\sqrt3+1}2=2$; $\theta=\frac{2\pi}3$ gives $-\frac{\sqrt3-1}2+\frac{3+\sqrt3}2=2$. Equivalently, $\theta=2n\pi\pm\frac\pi4+\frac{5\pi}{12}$ (the form you get by writing the left side as $r\cos(\theta-\alpha’)$ with $\alpha’=\frac{5\pi}{12}$).

The printed key gives $2n\pi\pm\frac\pi4+\frac\pi{12}$, which is wrong. Its $+$ branch gives $\theta=\frac\pi3$, and then the left side is $\frac{\sqrt3-1}2+\frac{3+\sqrt3}2=\sqrt3+1\neq2$. The hint’s $\alpha=\frac\pi{12}$ belongs to the sine form $\sin(\theta+\alpha)$ used above, whose family is $n\pi+(-1)^n\frac\pi4-\frac\pi{12}$; combining it with a $\pm\frac\pi4$ cosine-style family mixes the two forms.

Common mistakes

• Question 20: expanding $\cos2(\alpha-\beta)$ in terms of $\alpha$ and $\beta$ separately, which brings in $\theta$-dependent terms that never cancel cleanly. Writing the angle difference as $(\theta+\alpha)-(\theta+\beta)$ is what lets $a$ and $b$ appear.
• Question 21: dividing by $\cos\theta\cos\phi$ before collecting terms and losing the factor $(1-m)$ or $(1+m)$. Collect first, divide last.
• Question 22: forgetting that $\sin\left(\frac{3\pi}2-\alpha\right)=-\cos\alpha$ is negative. The even powers hide the sign here, but students who treat it as $+\cos\alpha$ often make the same slip in questions where the power is odd.
• Question 23: taking $\alpha$ and $\beta$ as angles with $\tan\alpha=\tan\beta$. The two roots must be genuinely different values of $\theta$ modulo $\pi$ for $\tan\alpha$ and $\tan\beta$ to be the two roots of the quadratic.
• Question 26: treating $\cos^4\frac{5\pi}8$ as negative. A fourth power of a negative cosine is positive, which is why the four terms pair off into two equal pairs.
• Questions 28 and 29: trusting the printed answer. In Question 28 the correct family is $\frac{n\pi}2+\frac\pi8$, not $\pm$. In Question 29 the shift is $\frac{5\pi}{12}$ in the $\pm\frac\pi4$ form; students who combine the hint’s $\alpha=\frac\pi{12}$ with a cosine form get a family that fails when substituted back. Always substitute one solution to check.

Practise next

• The same unit’s other exemplar pages: Short Answer Type, questions 1 to 19, Objective Type, questions 30 to 44, Objective Type, questions 45 to 59 and Fill in the Blanks, True or False and Matching, questions 60 to 76.
• NCERT textbook: Exercise 3.3, Exercise 3.2 and the Miscellaneous Exercise.