**Evaluating Trigonometric Ratios for Special Angles :**

Here we are going to see some example problems to evaluate trigonometric ratios for special angles.

**Question 1 :**

Verify the following equalities:

(i) sin^{2} 60°+ cos^{2} 60° = 1

**Solution :**

L.H.S :

= sin^{2} 60°+ cos^{2} 60°

= (√3/2)^{2} + (1/2)^{2}

= (3/4) + (1/4)

= (3 + 1)/4

= 4/ 4 = 1

Hence proved.

(ii) 1 + tan^{2} 30° = sec^{2} 30°

**Solution :**

L.H.S :

= 1 + tan^{2} 30°

= 1 + (1/√3)^{2}

= 1 + (1/3)

= 4/3

R.H.S :

= sec^{2} 30°

= (2/√3)^{2}

= 4/3

Hence proved.

(iii) cos 90° = 1 - 2 sin^{2} 45° = 2 cos^{2} 45° - 1

**Solution :**

**Part 1 :**

** = cos 90°**

** = 0 -------(1)**

**Part 2 :**

** = **1 - 2 sin^{2} 45°

** = 1 - 2(1/**√2)^{2}** **

** = 1 - 2(1/2**)** **

**= 0 ****-------(2)**

**Part 3 :**

** = ** 2 cos^{2} 45° - 1

** = 2(1/**√2)^{2}** - 1**

** = 2(1/2**)** - 1**

**= 0 ****-------(3)**

(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°

L.H.S :

= sin 30° cos 60° + cos 30° sin 60°

= (1/2) (1/2) + (√3/2) (√3/2)

= (1/4) + (3/4)

= (1 + 3)/4

= 4/4

= 1

R.H.S :

= sin 90°

= 1

Hence proved.

**Question 2 :**

Find the value of the following :

(tan 45°/cosec 30°)+(sec 60°/cot 45°)-(5 sin 90°/2cos 0°)

**Solution :**

= (tan45°/cosec30°)+(sec60°/cot45°)-(5sin90°/2cos0°)

tan45° = 1, cosec 30° = 2, sec 60° = 2, cot 45 = 1, sin 90 = 1 and cos 0 = 1

= (1/2) + (2/1) - (5(1)/2(1))

= (1/2) + (2/1) - (5/2)

= (1 + 2 - 5)/2

= -2/2

= -1

(ii) (sin 90°+cos 60°+cos 45°) × (sin 30°+cos 0°-cos 45°)

**Solution :**

= (sin 90°+cos 60°+cos 45°) × (sin 30°+cos 0°-cos 45°)

= [1 + (1/2) + (1/√2)] × [(1/2) + 1 - (1/√2)]

= [(3/2) + (1/√2)] × [(3/2) - (1/√2)]

= [(3/2)^{2} - (1/√2)^{2}]

= [(9/4) - (1/2)]

= (9 - 2)/4

= 7/4

(iii) sin^{2} 30° - 2 cos^{3} 60° + 3 tan^{4} 45°

**Solution :**

** = sin ^{2} 30° - 2 cos^{3} 60° + 3 tan^{4} 45°**

** = (1/2) ^{2} - 2(1/2)^{3} + 3(1)^{4}**

** = (1/4) - (1/4) + 3**

** = 3**

**Question 3 :**

Verify cos 3A = 4 cos^{3} A - 3 cosA , when A = 30°

**Solution :**

cos 3A = 4 cos^{3} A - 3 cosA

L.H.S :

= cos 3A

= cos 3(30)

= cos 90

= 0 ---(1)

R.H.S :

= 4 cos^{3} A - 3 cosA

= 4 cos^{3} 30 - 3 cos 30

= 4 (√3/2)^{3} - 3 (√3/2)

= (12√3/8) - (3√3/2)

= (3√3/2) - (3√3/2)

= 0 -----(2)

(1) = (2)

Hence proved.

**Question 4 :**

Find the value of 8 sin 2x cos 4x sin6x, when x = 15°

**Solution :**

** = 8 sin 2x cos 4x sin6x**

** = 8 **sin 2(15) cos 4(15) sin 6(15)

** = 8 **sin 30 cos 60 sin 90

= 8(1/2) (1/2) (1)

= 2

After having gone through the stuff given above, we hope that the students would have understood, "Evaluating Trigonometric Ratios for Special Angles"

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