Trigonometric Ratios Of Some Specific Angles
Trigonometric ratios of some common angles such as 0°, 45°, 90°, etc. are used very often in solving trigonometric questions.
Example 1: Find the value of cos 2A if A = 30°. Solution:
It is given that A = 30°
∴ cos 2A = cos (2 × 30°)
= cos 60°
=
Example 2: Prove that sin 2A = 2 sin A cos A, for A = 45°.
Solution:
L.H.S = sin 2 A = sin (2 × 45°)
= sin 90 °
= 1
R.H.S. = 2 sin A cos A = 2 × sin 45° × cos 45°
= 2 ×
= 1
L.H.S = R.H.S.
∴ sin 2A = 2 sin A cos A
Example 3: From the given figure XYZ, find ∠YZX and ∠ZXY.
Solution:
It is given that XY = and YZ = 3
Now,
∴ ∠YZX = 30°
By angle sum property in ΔXYZ, we obtain
∠ZXY = 180° − 30° − 90°
∴ ∠ZXY = 60°
Example 4: If sin (A + B) = 1 and tan (A −B) = 0, where 0°< A + B ≤ 90°, then find A and B.
Solution:
sin (A + B) = 1,
We know that sin 90° = 1
⇒ A + B = 90° ... (1) tan (A −B) = 0
And, we know that tan 0° = 0
⇒ A − B = 0° ... (2)
On adding (1) and (2), we obtain 2A = 90°
⇒ A = 45°
On putting this value of A in (1), we obtain B = 45°
Thus, the value of both A and B is 45°.
Example 5: Find the value of , if A = 60° and B = 30°.
Solution:
=
=
Example 6: Find the value of sin2 60° + 2 cos2 30° – tan2 45° + sec2 30°
Solution:
sin2 60° + 2 cos2 30° – tan2 45° + sec2 30°
Example 7: If sin x = sin 60° cos 30° + cos 60° sin 30° then find the value of x.
Solution:
We have
sin x = sin 60° cos 30° + cos 60° sin 30°
⇒ sin x = 1
⇒ sin x = sin 90°
⇒ x = 90°
Example 8: ABCD is a trapezium such that AB || CD and AD = BC.
Find the area of the trapezium.
Solution:
In right-angled ΔAED, we have
In ΔAED and ΔBFC, we have
AD = BC
∠AED = ∠BEC = 90° and
AE = BF (Perpendiculars drawn between two parallel lines)
∴ΔAED ΔBFC
By CPCT, we have
ED = FC = 83√83cm
ABFE is a rectangle, so we have AB = FE = 12 cm
Now, DC = ED + FE + FC = + 12 cm + =
Area of trapezium ABCD = (AB + DC) × AE