Advertisements
Advertisements
Question
Find the value of x, if sin 2x = 2 sin 45° cos 45°
Advertisements
Solution
sin 2x = 2 sin 45° cos 45°
sin 2x = `2(1/sqrt2)(1/sqrt2)`
sin 2x = 1 = sin 90°
2x = 90°
Hence, x = 45°
APPEARS IN
RELATED QUESTIONS
If the angle θ= –60º, find the value of cosθ.
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Solve.
sin42° sin48° - cos42° cos48°
Evaluate.
`(sin77^@/cos13^@)^2+(cos77^@/sin13^@)-2cos^2 45^@`
Write the maximum and minimum values of sin θ.
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
The value of cos2 17° − sin2 73° is
Prove that:
\[\frac{sin\theta \cos(90° - \theta)cos\theta}{\sin(90° - \theta)} + \frac{cos\theta \sin(90° - \theta)sin\theta}{\cos(90° - \theta)}\]
