Question

Topic :- Trigonometry

1) Let θ and Ф are the acute angles sinθ = 1/2 and cosФ= 1/3 Find the interval of θ+ Ф

Hint :- Answer is (π/2, 2π/3)

__________________________
2) sin²θ = (x+y)²/4xy Possible only, when

Hint :- Answer should be b,c

Note :- Options are in attachment

​

Answer 1

EXPLANATION.

Question = 1.

Let, θ and Φ are the acute angles.

⇒ sinθ = 1/2.

⇒ cosΦ = 1/3.

As we know that,

⇒ sinθ = 1/2 = π/6.

We can write equation as,

⇒ [π - π/6] = 5π/6.

⇒ sinθ = 1/2 = π/6 Or 5π/6.

⇒ cosФ = 1/3.

⇒ cosФ = π/3.

We can write equation as,

⇒ cos(π/3) > cosФ < cos(π/2).

⇒ π/3 > Ф > π/2.

Add θ in the equation, we get.

⇒ π/3 + π/6 < Ф + θ < π/2 + π/6.

⇒ (2π + π)/6 < Ф + θ < (3π + π)/6.

⇒ (3π/6) < Ф + θ < (4π/6).

⇒ π/2 < Ф + θ < 2π/3.

Answer = (π/2, 2π/3).

Question = 2.

sin²θ = (x + y)²/4xy.

As we know that,

Range of sinθ = [-1,1].

0 ≤ sin²θ ≤ 1.

⇒ 0 ≤ (x + y)²/4xy ≤ 1.

⇒ (x + y)² ≤ 4xy.

⇒ x² + y² + 2xy ≤ 4xy.

⇒ x² + y² - 2xy ≤ 0.

⇒ (x - y)² ≤ 0.

⇒ x = y. - - - - - (1).

⇒ 0 ≤ (x + y)²/4xy.

⇒ (x + y)² ≤ 0.

⇒ x² + y² + 2xy ≤ 0.

⇒ (x + y)² ≤ 0.

⇒ x = - y. - - - - - (2).

If you put x = y = 0 in the equation, we get.

⇒ sin²θ = (x + y)²/4xy.

⇒ sin²θ = (0)²/(0).

It is not defined.

⇒ |x| = |y| ≠ 0.

Answer 2

1) Let θ and Ф are the acute angles sinθ = 1/2 and cosФ= 1/3 Find the interval of θ+ Ф

Answer:-

Given: sinθ = 1/2 = sin (π/6) => as "θ" is acute angle.

$\fbox\pink{=>θ = π/6}$

Similarly cosФ = 1/3 = 0.333

cos60° = cos π/3 = 1/2 = 0.5

and cos90° = cos π/2 = 0

clearly 0.5 > 0.333 > 0

$\fbox\pink{=> π/3 < Ф < π/2}$

Now, θ+ Ф = π/6 + π/3 = π/2 ; π/6 + π/2 = 2π/3

$\fbox\pink{ = > π/2 < θ+ Ф < 2π/3 }$

Hence, Proved.

2) sin²θ = (x+y)²/4xy Possible only, when

Answer:-

$sin²θ = \frac{(x + y) {}^{2} }{4xy}$

Range of sin²θ [ 0,1 ]

$0 \leqslant \frac{(x + y) {}^{2} }{4xy} \leqslant 1$

$0 \leqslant \frac{x {}^{2} + y {}^{2} + 2xy }{4xy} \leqslant 1$

This condition is satisfy only if x = y

but x < 0 , y < 0 and x > 0 , y > 0 both are possible.