Problems on Trigonometric Ratio of Standard Angle

How to solve the problems on Trigonometric Ratio of Standard Angle?

We know the standard angles are 0°, 30°, 45°, 60° and 90°. The questions are based on these standard angles. Here we will learn how to solve the standard angle of trigonometry related question.

Standard angles in trigonometry generally mean those angles whose trigonometric ratios can determine without using calculators. To find the values of trigonometric ratios of these standard angles we need to follow the trigonometric table.


Worked-out problems on trigonometric ratio of standard angle:

1. If β = 30°, prove that 3 sin β - 4 sin\(^{3}\) β = sin 3β.

Solution:

L.H.S = 3 sin β - 4 sin\(^{3}\) β

 = 3 sin 30° – 4. sin\(^{3}\) 30°

= 3 ∙ (1/2) - 4 ∙ (1/2)\(^{3}\)

= 3/2 – 4 ∙  1/8

3/2 – ½

=  1

R.H.S. = sin 3A

= sin 3 ∙ 30°

= sin 90°

= 1

Therefore, L.H.S. = R.H.S. (Proved)


2. Find the value of 4/3 tan\(^{2}\) 60° + 3 cos\(^{2}\)  30° - 2 sec\(^{2}\)  30° - 3/4 cot\(^{2}\)  60°

Solution:

The given expression

\(\frac{4}{3} \cdot (\sqrt{3})^{2} + 3 \cdot  (\frac{\sqrt{3}}{2})^{2} - 2  \cdot  (\frac{2\sqrt{3}}{3})^{2} - \frac{3}{4} \cdot  (\frac{\sqrt{3}}{3})^{2}\)

= \(\frac{4}{3} \cdot  3 + 3 \cdot  \frac{3}{4} - 2 \cdot  \frac{12}{9} - \frac{3}{4} \cdot  \frac{3}{9}\)

= 4 + 9/4 - 8/3 – 1/4

= 10/3

= \(3\tfrac{1}{3}\)

 

3. If θ = 30°, prove that cos 2θ = cos\(^{2}\) θ -  sin\(^{2}\) θ

Solution:

L. H. S. = cos 2θ

= cos 2 ∙ 30°

= cos 60°

=  1/2

And R. H. S. = cos\(^{2}\) θ -  sin\(^{2}\) θ

= cos\(^{2}\) 30° - sin\(^{2}\) 30°

= (√3/2)\(^{2}\) – (1/2)\(^{2}\)

= ¾ - ¼

= 1/2

Therefore,  L.H.S = R.H.S. (Proved)


4. If A = 60° and B = 30°, verify that sin (A - B) = sin A cos B - cos A sin B

Solution:

L.H.S. = sin (A - B)

= sin (60° - 30°)

= sin 30°

= ½

R.H.S. = sin A cos B - cos A sin B

= sin 60° cos 30° - cos 60° sin 30°

= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} - \frac{1}{2} \times \frac{1}{2}\)

= ¾ - ¼

= 2/4

= ½

Therefore, L.H.S. = R.H.S. (Proved)


5. If sin (x + y) = 1 and cos (x - y) = \(\frac{\sqrt{3}}{2}\), find x and y.

Solution:

sin (x + y) = 1

 sin (x + y) = sin 90°, [since sin 90° = 1]

⇒ x + y = 90° .........................(A)

cos (x - y) = \(\frac{\sqrt{3}}{2}\)

⇒ cos (x - y) = cos 30°

⇒ x - y = 30° .........................(B)

Adding, (A) and (B), we get

                   x + y =  90°

                   x - y =  30°

                  2x     = 120°

                    x = 60°, [Dividing both sides by 2]

Putting the value of x = 60° in (A) we get,

60° + y = 90°

Subtract 60° from both sides

                    60° + y = 90°

                   -60°       -60°

                            y = 30°

Therefore, x = 60° and y = 30°.

 Trigonometric Functions





11 and 12 Grade Math

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