JEE Advanced (Subjective Type Questions): Trigonometric Functions & Equations | Chapter-wise Tests for JEE Main & Advanced PDF Download

Clear all your JEE doubts with EduRev Join for Free

Join for Free

Q. 1. If  and  find the possible values of (α + β).    (1978)

Ans. 

 Sol. We know tan

2. (a) Draw the graph of y =(sinx + cosx) from x = to

(b) If cos and α, β lies between 0 and  , find tan2α. (1979)

Ans. 

Sol.  (a) Given: (sinx + cosx) sin   ...(1)

Now, to draw the graph of we first draw

the graph of y = sin x and then on shifting it by   we will obtain the required graph as shown in figure given below.

(b)

∴   tan 2α = tan [(α + β) + (α - β) ]

3. Given α + β - γ = π, prove that   (1980)

Ans. 

Sol.  Given α + β – γ = π and to prove that

sin² α + sin²β – sin²γ = 2sinα sinβ cosγ

L.H.S. = sin²α + sin²β – sin²γ

[Using sin²α – sin²β = sin (A + B) sin (A – B)]

= sin²α + sin(β + γ) sin (β – γ)

= sin²α + sin(β + γ) sin (p – α)         (∵ α + β – γ = π)

= sin²α + sin(β + γ) sinα

= sinα (sinα + sin(β + γ))

= sinα [sin [π – (β - γ)] + sin(β + γ)]

= sinα [sin(β - γ) + sin(β + γ)]

= sinα [2 sinβ cosγ] = 2 sinα sinβ cosγ =  R.H.S.

4. Given  and f (x) = cos x – x (1 + x); find f (A). (1980)

Ans.  

Sol. 

f (x) = cos x – x (1+ x)

f '(x) = –  sin x –  1– 2x < 0, ∀ x∈A

∴ f is a decreasing function.

5. For all θ in [0, π/ 2] show that, cos (sinθ) ≥ sin (cos q) . (1981 - 4 Marks)

Ans.  Sol. We have

cosθ + sinθ

∴ cosθ + sinθ <π/ 2 ⇒ cosθ < π/ 2 - sinθ ...(1)

As  q ∈ [0,π / 2] in which sinθ increases.

∴ Taking sin on both sides of eq. (1), we get

sin (cosθ ) < sin ( π /2 –sinθ )

sin (cosθ ) < cos (sinθ )

⇒ cos (sinθ ) > sin (cosθ ) ....(1)

Hence the result.

6. Without using tables, prove that

(sin 12°) (sin 48°) (sin 54°)  =  (1982 - 2 Marks)

Ans.  Sol.  L.H.S. = sin 12° sin 48° sin 54° =  [2 sin 12° cos 42°] sin 54°

Now we know that sin 54°

∴ We get, =

7. Show that 16cos  (1983 - 2 Marks)

Ans.  Sol.  We know that, 

cos A cos 2A cos 4A  . .  . . cos 2ⁿ
 

      (where A = 2π/15)

8. Find all the solution of 4 cos ² x sin x - 2 sin² x= 3 sinx           (1983 - 2 Marks)

Ans.  Sol. Given eq. is, 

4 cos² x sin x – 2 sin² x = 3 sin  x

⇒ 4 cos² x sin x – 2 sin² x – 3 sin x = 0

⇒ 4 (1 – sin² x) sin x – 2 sin² x – 3 sin x = 0

⇒ sin x [ 4 sin² x + 2 sin x – 1] = 0

⇒ either sin x = 0 or 4 sin² x + 2 sin x – 1 = 0

If sin x = 0 ⇒ x = np

⇒ If 4 sin² x + 2 sin x – 1 = 0 ⇒ sin x =

If  sin x =

then

If sin x == sin(-54⁰)  

then x  

Hence,  

where n is some integer

9. Find the values of x∈(–π, +π) which satisfy the equation

      (1984 - 2 Marks)

Ans.  Sol. The given equation is

               NOTE THIS STEP

The values of   

10. Prove that tanα + 2 tan 2α + 4 tan 4α + 8 cot 8α = cotα        (1988 - 2 Marks)

Ans.  Sol. We know that tan

 ⇒ cot α – tan α = 2 cot 2α

Now we have to prove

tanα + 2 tan 2α + 4 tan 4α + 8 cot 8α = cot α

LHS

tan α + 2 tan 2α + 4 tan 4α + 4 (2 cot 2 . 4α)

= tan α + 2 tan 2α + 4 tan 4α + 4 (cot 4α - tan 4α) [Using (1)]

= tan α + 2 tan 2α + 4 tan 4α + 4 cot 4α - 4 tan 4α

= tan α + 2 tan 2α + 2 (2 cot 2. 2  a )

= tan α + 2 tan 2α + 2 (cot α –  tan 2α)

= tan α + 2 tan 2α + 2 (2 cot 2α-tan 2α ) [Using (1)]

= tan α + 2 cot 2α

= tan α + (cot α-tan α) [Using (1)]

= cot α = RHS.

11. ABC is a triangle such that sin(2A + B) = sin (C – A) = – sin (B + 2C) =

If A, B and C are in arithmetic progression, determine the values of A, B and C. (1990 -  5 Marks)

Ans.  Sol. Given that in ΔABC, A, B and C are in A.P.

∴ A + C = 2B

also A + B + C = 180° ⇒ B + 2B = 180° ⇒ B = 60°

Also given that, sin (2A + B) = sin (C – A) = – sin (B + 2C) = 

⇒ sin (2A + 60°) = sin (C – A) = – sin (60 + 2C) =      ..(1)

From eq. (1), we have

sin (2A + 60°) =      ⇒ 2A + 60° = 30° ,  150°

but A can not be –ve

∴ 2A + 60° = 150°   ⇒   2A = 90 °  ⇒ A = 45°

Again from (1)  sin (60° + 2C) = -

⇒ 60° + 2C = 210°    or     330°
⇒ C = 75°                 or     135°

Also from (1)  sin (C – A) =⇒ C –  A = 30°, 150°

For   A = 45°, C =75°   or 195° (not possible)    ∴ C = 75°

Hence we have A = 45° , B = 60°, C = 75°

12. If exp {(sin²x + sin⁴x + sin⁶x +   ............... ∞ ) In 2} satisfies the equation x²– 9x + 8 = 0, find the value of

    (1991 -  4 Marks)

Ans.  Sol. Let  y = exp [sin² x +sin⁴ x +sin⁶ x + . . . . ∞ ] ln 2

As y satisfies the eq.

x² – 9x + 8 = 0          ∴ y² – 9y + 8 = 0

⇒ (y – 1)  (y – 8) = 0 ⇒ y = 1, 8

⇒ tan 2 x = 0   or    tan 2x = 3

⇒ tan x = 0    or

⇒ x = 0    or    x = π/3,   2 π/3

But given that 0  < x < π/2 ⇒ x = π/3

Hence  

13. Show that the value of  wherever defined never lies between   and 3. (1992 -  4 Marks)

Ans.  

Sol. 

Let  

⇒ 3y – 3 tan² x = 1– 3 tan² x

⇒ (y – 3) tan² x = 3y – 1 ⇒

(L.H.S. being a prefect square)

⇒⇒ (3y - 1) (y - 3)>0

Thus y never lies between  and 3 

14. Determine the smallest positive value of x (in degrees) for which tan(x + 100°) = tan (x + 50°) tan(x) tan (x – 50°).      (1993 -  5 Marks)

Ans.  Sol. Given that, tan (x + 100°) = tan (x + 50°) tan x tan (x – 50°)

⇒ tan (x + 50°) tan (x – 50°)

⇒

⇒

Applying componendo and dividendo, we get

⇒

⇒ 2 sin (2x + 100°) cos 2x = – 2 sin 100° cos 100°

⇒ sin (4x + 100°) + sin 100° =  – sin 200°

⇒ sin (4x + 10° + 90°) + sin (90° + 10°) = – sin (180 + 20°)

⇒ cos (4x + 10°) + cos 10° =  sin 20°

⇒ cos (4x + 10°) = sin 20° – cos 10°

⇒ cos (4x + 10°) = sin 20° – sin 80°

= – 2 cos 50° sin 30° = – 2 cos 50°. = –cos 50° = cos 130°

⇒ 4x  + 10° = 130° ⇒ x = 30° 

15. Find the smallest positive number p for which the equation cos(p sin x) = sin(p cos x) has a solution x ∈ [0,2π].   (1995 -  5 Marks)

Ans. 

 Sol. Given that cosθ = sin φ

where  θ = p sin x,  φ = p cos x

Above is possible when both   or

∴ p sin x =   or       p sin x =

and p cos x =      or p cos x =

Squaring and adding,

only for least positive value  or  

16. Find all values of θ in the interval  satisfying the equation      (1996 - 2 Marks)

Ans.  

Sol. Given  : 

or

Let us put tan² θ = t

∴ (1– t) ( 1 + t) + 2^t = 0 or     1– t² + 2^t = 0

It is clearly satisfied by t = 3.

as – 8 + 8 = 0   ∴     tan² θ = 3

∴ p = ± π/3 in the given interval.

17. Prove that the values of the function    do not lie between  and 3 for any real x. (1997 - 5 Marks)

Ans.  

Sol.  Let

We have

(the expression is not defined if tan x = 0)

⇒ 3y – (tan² x) y = 1– 3 tan² x ⇒ 3y – 1=  (y – 3) tan² x

⇒  

Since tan² x > 0,  we get      (3y – 1)  (y – 3) > 0

⇒      or      y > 3

This shows that y cannot lie between  and  3.

18. Prove that   where n ≥ 3 is an integer. (1997 - 5 Marks)

Ans. 

 Sol. Expanding the sigma on putting k = 1,  2,  3, ......, n

S = (n – 1)  

+ 1.cos (n – 1) ... (1)

We know that cosθ = cos (2π – θ)

Replacing each angle θ by 2π – θ in (1),  we get

S = (n – 1) cos (n – 1)  + (n – 2) cos (n – 2)  + ......+ 1.cos    by   (1) .....(2)

Add terms in (1) and (2) having the same angle and take n common

∴ 2S=  

Angles are in A.P. of  

        NOTE THIS STEP

 = n .1 cosπ = – n ∵ sin (π – θ) = sinθ ∴  S = – n/2 

19. In any triangle ABC, prove that (2000 - 3 Marks)

Ans.  Sol. We have, A + B + C = π

⇒  

or

⇒  

⇒  

20. Find the range of values of t for which   

(2005 - 2 Marks)

Ans.  Sol. Given that, 2 sin t t ∈ [-π/ 2,π/ 2]

This can be written as

(6 sin t – 5) x² + 2 (1– 2 sin t) x – (1+ 2 sin t) = 0

For given equation to hold, x should be some real number, therefore above equation should have real roots i.e., D ≥ 0

⇒ 4 (1– 2 sin t)² + 4 (6 sin t – 5) (1+ 2 sin t) ≥ 0

⇒

 or  

 or  

(Note that sin x is an increasing function from –π/2 to π/2)

∴ range of  t is