Advertisements
Advertisements
प्रश्न
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A – 1 = 0
योग
उत्तर
2 cos2 A – 1 = 0
`=> cos^2A = 1/2`
`=> cosA = 1/sqrt(2)`
We know `cos 45^@ = 1/sqrt(2)`
Hence, A = 45°
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Evaluate `(sin 18^@)/(cos 72^@)`
Evaluate.
`cos^2 26^@+cos65^@sin26^@+tan36^@/cot54^@`
Find the value of x, if sin x = sin 60° cos 30° + cos 60° sin 30°
Use tables to find sine of 34° 42'
Find A, if 0° ≤ A ≤ 90° and cos2 A – cos A = 0
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
If tanθ = 2, find the values of other trigonometric ratios.
Write the maximum and minimum values of sin θ.
What is the maximum value of \[\frac{1}{\sec \theta}\]
If tan θ = 1, then sin θ . cos θ = ?
