NCERT Exemplar Solutions Class 11 Maths Trigonometric Functions — Fill in the Blanks, True or False and Matching Q60–76

NCERT Exemplar Problems, Class 11 Mathematics — Trigonometric Functions, Fill in the Blanks (60 to 67), True or False (68 to 75) and Matching (76) questions 60 to 76. All 17 questions solved.

Seventeen short-form questions. The blanks (60 to 67) ask for a value, an expression, a quadratic or a range; the True or False statements (68 to 75) need a justification, not just a verdict; and Question 76 matches four product and shift identities.

One answer in the printed key is wrong, Question 73, and Question 72 is misprinted (the “$=0$” is missing). Both are explained where they occur.

Key insight. For a True or False question, a statement is false as soon as one case fails, so test a boundary case first. Here that decides 69, 70, 72 and 73: the extreme value of $\sin A+\sin2A+\sin3A$, the sine and cosine of $10^\circ$, the range of $\sin^4\theta-2\sin^2\theta-1$, and the point $x=2n\pi$ where $\mathrm{cosec}\,x$ is undefined.

Question 60

Fill in the blank. The value of $\dfrac{\sin50^\circ}{\sin130^\circ}$ is ____.

Solution. $\sin130^\circ=\sin(180^\circ-50^\circ)=\sin50^\circ$, so the ratio is $1$.

Question 61

Fill in the blank. If $k=\sin\left(\dfrac\pi{18}\right)\sin\left(\dfrac{5\pi}{18}\right)\sin\left(\dfrac{7\pi}{18}\right)$, then the numerical value of $k$ is ____.

Solution. The angles are $10^\circ$, $50^\circ$, $70^\circ$. They have the pattern $\theta,\ 60^\circ-\theta,\ 60^\circ+\theta$ with $\theta=10^\circ$, and for that pattern $\sin\theta\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\frac14\sin3\theta$. (To see it, $\sin(60^\circ-\theta)\sin(60^\circ+\theta)=\sin^260^\circ-\sin^2\theta=\frac34-\sin^2\theta$, and $\sin\theta\left(\frac34-\sin^2\theta\right)=\frac14(3\sin\theta-4\sin^3\theta)=\frac14\sin3\theta$.) So $$k=\frac14\sin30^\circ=\frac14\cdot\frac12=\frac18.$$

Question 62

Fill in the blank. If $\tan A=\dfrac{1-\cos B}{\sin B}$, then $\tan2A=$ ____.

Solution. The half-angle identity gives $\dfrac{1-\cos B}{\sin B}=\dfrac{2\sin^2\frac B2}{2\sin\frac B2\cos\frac B2}=\tan\dfrac B2$. So $\tan A=\tan\frac B2$, $A=\frac B2$ (plus a multiple of $\pi$), and $\tan2A=\tan B$.

Question 63

Fill in the blanks. If $\sin x+\cos x=a$, then (i) $\sin^6x+\cos^6x=$ ________ (ii) $|\sin x-\cos x|=$ ________.

Solution. Square the given: $1+2\sin x\cos x=a^2$, so $\sin x\cos x=\dfrac{a^2-1}2$.

(i) $\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$. Substituting, $$1-3\cdot\frac{(a^2-1)^2}4=\frac14\left[4-3(a^2-1)^2\right].$$

(ii) $(\sin x-\cos x)^2=1-2\sin x\cos x=1-(a^2-1)=2-a^2$, so $|\sin x-\cos x|=\sqrt{2-a^2}$.

Question 64

Fill in the blank. In a triangle $ABC$ with $\angle C=90^\circ$ the equation whose roots are $\tan A$ and $\tan B$ is ____.

Solution. With $C=90^\circ$, $A+B=90^\circ$, so $B=90^\circ-A$ and $\tan B=\cot A$. Then the product of the roots is $\tan A\tan B=1$, and the sum is $$\tan A+\cot A=\frac{1}{\sin A\cos A}=\frac2{\sin2A}.$$ A quadratic with sum $S$ and product $P$ is $x^2-Sx+P=0$.

Question 65

Fill in the blank. $3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin^6x+\cos^6x)=$ ____.

Solution. Let $p=\sin x\cos x$. Then $(\sin x-\cos x)^2=1-2p$, $(\sin x+\cos x)^2=1+2p$, and $\sin^6x+\cos^6x=1-3p^2$. So the expression is $$3(1-2p)^2+6(1+2p)+4(1-3p^2)=3-12p+12p^2+6+12p+4-12p^2=13.$$ All the $p$ terms cancel, so the value is a constant.

Question 66

Fill in the blank. Given $x>0$, the values of $f(x)=-3\cos\sqrt{3+x+x^2}$ lie in the interval ____.

Solution. The inner function $g(x)=\sqrt{3+x+x^2}$ is continuous and strictly increasing for $x>0$, and takes every value in $(\sqrt3,\infty)$. That interval is longer than a full period of cosine, so $\cos g(x)$ takes every value in $[-1,1]$ (including both $-1$ and $1$, at suitable large $x$). Multiplying by $-3$ gives every value in $[-3,3]$. A numerical scan of $f$ over $0<x<100$ gives a minimum of $-3.000$ and a maximum of $3.000$.

Question 67

Fill in the blank. The maximum distance of a point on the graph of the function $y=\sqrt3\sin x+\cos x$ from the $x$-axis is ____.

Solution. The distance of the point $(x,y)$ from the $x$-axis is $|y|$. Write $\sqrt3\sin x+\cos x=2\sin\left(x+\frac\pi6\right)$, since $\sqrt{(\sqrt3)^2+1^2}=2$. Its absolute value has maximum $2$.

Question 68

State whether the statement is True or False, and justify. If $\tan A=\dfrac{1-\cos B}{\sin B}$, then $\tan2A=\tan B$.

Solution. As in Question 62, $\dfrac{1-\cos B}{\sin B}=\tan\frac B2$, so $\tan A=\tan\frac B2$ and $A=\frac B2+n\pi$. Then $2A=B+2n\pi$ and $\tan2A=\tan B$.

Question 69

State whether the statement is True or False, and justify. The equality $\sin A+\sin2A+\sin3A=3$ holds for some real value of $A$.

Solution. Each sine is at most $1$, so the sum can reach $3$ only if $\sin A=\sin2A=\sin3A=1$ at the same $A$. But $\sin A=1$ forces $A=\frac\pi2+2n\pi$, and then $\sin2A=\sin(\pi+4n\pi)=0\ne1$. So the sum is always strictly less than $3$.

Question 70

State whether the statement is True or False, and justify. $\sin10^\circ$ is greater than $\cos10^\circ$.

Solution. $\cos10^\circ=\sin80^\circ$, and on $[0^\circ,90^\circ]$ sine increases, so $\sin10^\circ<\sin80^\circ=\cos10^\circ$. Numerically $\sin10^\circ\approx0.174$ and $\cos10^\circ\approx0.985$.

Question 71

State whether the statement is True or False, and justify. $\cos\dfrac{2\pi}{15}\cos\dfrac{4\pi}{15}\cos\dfrac{8\pi}{15}\cos\dfrac{16\pi}{15}=\dfrac1{16}$.

Solution. Let $x=\frac{2\pi}{15}$. The four angles are $x,2x,4x,8x$, and the repeated-doubling identity gives $\cos x\cos2x\cos4x\cos8x=\dfrac{\sin16x}{16\sin x}$ (multiply by $\sin x$ and apply $\sin2\theta=2\sin\theta\cos\theta$ four times). Here $16x=\frac{32\pi}{15}=2\pi+\frac{2\pi}{15}$, so $\sin16x=\sin x$. The product is $\dfrac1{16}$.

Question 72

State whether the statement is True or False, and justify. One value of $\theta$ which satisfies the equation $\sin^4\theta-2\sin^2\theta-1\ (=0)$ lies between $0$ and $2\pi$.

The book prints this sentence without “$=0$”. Read as written, the expression is not an equation; the intended statement is $\sin^4\theta-2\sin^2\theta-1=0$, and that is what is solved here.

Syllabus note: solving trigonometric equations is no longer in the current NCERT Class 11 syllabus.

Solution. Let $u=\sin^2\theta\in[0,1]$. Then $u^2-2u-1=0$ gives $u=1\pm\sqrt2$. The root $1+\sqrt2\approx2.41$ exceeds $1$ and $1-\sqrt2\approx-0.41$ is negative, so neither is a possible value of $\sin^2\theta$. (Equivalently $u^2-2u-1=(u-1)^2-2\le-1$ on $[0,1]$, never $0$.) No real $\theta$ satisfies it, so none lies in $(0,2\pi)$.

Question 73

State whether the statement is True or False, and justify. If $\mathrm{cosec}\,x=1+\cot x$, then $x=2n\pi,\ 2n\pi+\dfrac\pi2$.

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

Solution. Write both sides over $\sin x$ (which must be non-zero): $\dfrac1{\sin x}=1+\dfrac{\cos x}{\sin x}$, so $1=\sin x+\cos x$. From Question 15 of this unit, that has solutions $x=2n\pi$ and $x=2n\pi+\frac\pi2$.

But the original equation requires $\sin x\ne0$. At $x=2n\pi$, $\sin x=0$, so $\mathrm{cosec}\,x$ and $\cot x$ do not exist. Only $x=2n\pi+\frac\pi2$ survives, where $\mathrm{cosec}\,x=1$ and $1+\cot x=1+0=1$. The statement claims $x=2n\pi$ is a solution, which is false.

The printed key says True; that is wrong for this reason.

Question 74

State whether the statement is True or False, and justify. If $\tan\theta+\tan2\theta+\sqrt3\tan\theta\tan2\theta=\sqrt3$, then $\theta=\dfrac{n\pi}3+\dfrac\pi9$.

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

Solution. Rearrange: $\tan\theta+\tan2\theta=\sqrt3(1-\tan\theta\tan2\theta)$, so $\dfrac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta}=\sqrt3$ (the denominator cannot be zero, since then $\tan\theta+\tan2\theta$ would also be zero and $\tan\theta\tan2\theta=1$ would be impossible alongside it). That is $\tan3\theta=\sqrt3=\tan\frac\pi3$, so $3\theta=n\pi+\frac\pi3$ and $\theta=\frac{n\pi}3+\frac\pi9$. Check: at $\theta=\frac\pi9$ the left side evaluates to $\sqrt3$ numerically.

Question 75

State whether the statement is True or False, and justify. If $\tan(\pi\cos\theta)=\cot(\pi\sin\theta)$, then $\cos\left(\theta-\dfrac\pi4\right)=\pm\dfrac1{2\sqrt2}$.

Syllabus note: solving trigonometric equations is no longer in the current NCERT Class 11 syllabus.

Solution. Write $\cot(\pi\sin\theta)=\tan\left(\frac\pi2-\pi\sin\theta\right)$. Equal tangents give $\pi\cos\theta=n\pi+\frac\pi2-\pi\sin\theta$, so $$\sin\theta+\cos\theta=n+\frac12=\frac{2n+1}2.$$ Since $\sin\theta+\cos\theta=\sqrt2\cos\left(\theta-\frac\pi4\right)$, we get $\cos\left(\theta-\frac\pi4\right)=\dfrac{2n+1}{2\sqrt2}$. This must lie in $[-1,1]$, so $|2n+1|\le2\sqrt2\approx2.83$. The odd integer $2n+1$ is therefore $\pm1$, giving $\pm\dfrac1{2\sqrt2}$. (A numerical root search over $[0,2\pi]$ finds only these two values, $\pm0.354$.)

Question 76

In the following, match each item given under the column $C_1$ to its correct answer given under the column $C_2$.

Solution. (a) Expand: $\sin(x+y)\sin(x-y)=(\sin x\cos y)^2-(\cos x\sin y)^2=\sin^2x(1-\sin^2y)-(1-\sin^2x)\sin^2y=\sin^2x-\sin^2y$. This is (iv).

(b) Similarly $\cos(x+y)\cos(x-y)=\cos^2x\cos^2y-\sin^2x\sin^2y=\cos^2x(1-\sin^2y)-(1-\cos^2x)\sin^2y=\cos^2x-\sin^2y$. This is (i).

(c) $\cot\left(\frac\pi4+\theta\right)=\dfrac1{\tan\left(\frac\pi4+\theta\right)}=\dfrac{1-\tan\theta}{1+\tan\theta}$, since $\tan\left(\frac\pi4+\theta\right)=\dfrac{1+\tan\theta}{1-\tan\theta}$. This is (ii).

(d) $\tan\left(\frac\pi4+\theta\right)=\dfrac{1+\tan\theta}{1-\tan\theta}$. This is (iii).

Common mistakes

• Question 61: trying to evaluate $\sin10^\circ\sin50^\circ\sin70^\circ$ with a calculator-style approach and getting a decimal. The three angles form $\theta,\ 60^\circ-\theta,\ 60^\circ+\theta$, which is the pattern for $\frac14\sin3\theta$.
• Question 63: leaving the answer to (ii) as $\sqrt{1-2\sin x\cos x}$ or dropping the absolute value. The question asks for $|\sin x-\cos x|$, which is exactly $\sqrt{2-a^2}$, a non-negative number.
• Question 66: writing the range as $[-3\cos\sqrt3,\,\infty)$ or similar from the starting value at $x=0$. The argument keeps growing without bound, so the cosine sweeps its whole range again and again.
• Question 69: adding the three maxima ($1+1+1=3$) and calling the statement true. The maxima occur at different angles, so they are never reached together.
• Question 72: reading the printed expression as already an equation and giving up. It is a misprint; once the “$=0$” is restored the argument via $\sin^2\theta\in[0,1]$ ends the question.
• Question 73: accepting both families from $\sin x+\cos x=1$. The step “multiply by $\sin x$” introduced $x=2n\pi$, where $\mathrm{cosec}\,x$ is undefined. This is the same trap as Question 19 on the Short Answer page.

Practise next

• The same unit’s other exemplar pages: Short Answer Type, questions 1 to 19, Long Answer Type, questions 20 to 29, Objective Type, questions 30 to 44 and Objective Type, questions 45 to 59.
• NCERT textbook: Exercise 3.3, Exercise 3.2 and the Miscellaneous Exercise.