Problems on Signs of Trigonometrical Ratios

We will learn how to solve various type of problems on signs of trigonometrical ratios of any angles.

1. For what real values of x is the equation 2 cos θ = x + 1/x possible?

Solution:

Given, 2 cos θ = x + 1/x

⇒ x\(^{2}\) - 2 cos θ ∙ x + 1 = 0, which is a quadratic in x. As x is real, distinct ≥ 0

⇒ (- 2 cos θ)\(^{2}\) - 4 ∙ 1 ∙ 1 ≥ 0

⇒ cos\(^{2}\) θ ≥ 1 but cos^2 θ ≤ 1

⇒ cos\(^{2}\) θ = 1

⇒ cos θ = 1, 1

Case I: When cos θ = 1, we get,

 x\(^{2}\) - 2x + 1 =0

⇒ x = 1

Case II: When cos θ = -1, we get,

x\(^{2}\) + 2x + 1 =0

⇒ x = -1.

Hence the values of x are 1 and -1.

 

2.  Solve sin θ + √3cos θ = 1, (0 < 0 < 360°).

Solution:

sin θ + √3cos θ = 1                       

⇒ √3cos θ = 1- sin θ  

⇒  (√3cos θ)\(^{2}\) = (1- sin θ)\(^{2}\)

⇒ 3cos\(^{2}\) θ = 1 - 2sin θ + sin\(^{2}\) θ

⇒ 3(1 - sin\(^{2}\) θ) - 1 + 2sin θ - sin\(^{2}\) θ = 0

⇒ 2 sin\(^{2}\) θ - sin θ - 1 = 0

⇒ 2 sin\(^{2}\) θ - 2 sin θ + sin θ - 1 = 0

⇒ (sin θ - 1)(2 sin θ +1  ) =0

Therefore, either sin θ - 1 = 0 or, 2 sin θ + 1 =0

If sin θ - 1= 0 then

sin θ = 1 = sin 90°                               

Therefore, θ = 90°

Again, 2 sin θ + 1 =0 gives, sin θ = -1/2

Now, since sin θ is negative, hence θ lies either in the third or in the fourth quadrant.

Since sin θ = -1/2 = - sin 30° = sin (180° + 30°) = sin 210°

and sin θ = - 1/2 = - sin 30° = sin (360° - 30°) = sin 330°

Therefore, θ = 210° or 330°

Therefore, the required solutions in

0 < θ < 360°are: 90°, 210° and 330°.

 

3. If the 5 sin x = 3, find the value of \(\frac{sec x  -  tan x}{sec x  +  tan x}\).

Solution:

Given 5 sin x = 3

⇒ sin x = 3/5.

Now \(\frac{sec x - tan x}{sec x + tan x}\)

 = \(\frac{\frac{1}{cos x}  -  \frac{sin x}{cos x}}{\frac{1}{cos x} + \frac{sin x}{cos x}}\)

= \(\frac{1  -  sin x}{1  +  sin x}\)

= \(\frac{1  -  \frac{3}{5}}{1  +  \frac{3}{5}}\)

= \(\frac{\frac{2}{5}}{\frac{8}{5}}\)

= 2/8

= ¼.

 

4. A, B, C, D are the four angles, taken in order of a cyclic quadrilateral. Prove that,
cot A + cot B + cot C + cot D = 0.

Solution:

We know that the opposite angles of a cyclic quadrilateral are supplementary.

Therefore, by question we have,

A + C= 180° or, C = 180° - A;

And B + D= 180° or, D = 180° - B.

Therefore, L. H. S. = cot A + cot B + cot C + cot D

= cot A + cot B + cot (180° - A) + cot (180° - B)

= cot A + cot B - cot A - cot B

= 0. Proved.

 

5. If tan α = - 2, find the values of the remaining trigonometric function of α.

Solution:

Given tan α = - 2 which is - ve, therefore, α lies in second or fourth quadrant.

Also sec\(^{2}\) α = 1 + tan\(^{2}\) α = 1 + (-2)\(^{2}\) = 5

⇒ sec α = ± √5.

Two cases arise:

Case I. When α lies in the second quadrant, sec α is (-ve).

Therefore, sec α = -√5

cos α = - 1/√5

sin α = \(\frac{sin \alpha}{cos \alpha} \cdot cos \alpha\) = tan α cos α = -2 -\(\frac{1}{\sqrt{5}}\) = 2/√5

⇒ csc α = √5/2.

Also tan α = -2

⇒ cot α = ½.

Case II. When α lies in the fourth quadrant, sec α is + ve

Therefore, sec α = √5

⇒ cos α = 1/√5

sin α = \(\frac{sin \alpha}{cos \alpha} \cdot cos \alpha\) = tan α cos α = -2 \(\frac{1}{\sqrt{5}}\) = 2/√5

 

6. If tan (α - β) = 1, sec (α + β) = 2/√3, find positive magnitudes of α and β.

Solution: 

We have, tan (α - β) = 1 = tan 45°                          

Therefore, α - β = 45° ………………. (1)

Again, sec (α + β)= 2/√3                 

⇒ cos (α + β)= √3/2 

⇒ cos (α + β) = cos 30°  or, cos (360° - 30°) = cos 330°  

Therefore, α + β = 30°  or, 330°

Since α and β are positive and α - β = 45°, hence we must have,

α + β = 330° …………….. (2)

(1)+ (2) gives, 2a = 375°            

⇒ α = {187\(\frac{1}{2}\)}°

and (2) - (1) gives,

2β = 285° or, β = {142\(\frac{1}{2}\)}°






11 and 12 Grade Math

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