Advertisements
Advertisements
प्रश्न
Find the value of x, if sin 2x = 2 sin 45° cos 45°
Advertisements
उत्तर
sin 2x = 2 sin 45° cos 45°
sin 2x = `2(1/sqrt2)(1/sqrt2)`
sin 2x = 1 = sin 90°
2x = 90°
Hence, x = 45°
संबंधित प्रश्न
What is the value of (cos2 67° – sin2 23°)?
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° – 3A) . cosec 42° = 1
Use tables to find sine of 34° 42'
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A + cos A – 1 = 0
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
