Using principal value, evaluate the following: sin-1 sin3π5
Prove that: sin-1 45 + sin-1 513 + sin-1 1665 = π2
Solve for x: tan-1 3x + tan-1 2x = π4
tan-1 3x + tan-1 2x = π4⇒ tan-1 5x1 - 6x2 = π4, 3x × 2x < 1⇒ tan tan-1 5x1 - 6x2 = tan π4⇒ 5x1 - 6x2 = 1⇒ 1 - 6x2 = 5x⇒ 6x2 + 5x - 1 = 0⇒ 6x2 + 6x - x - 1 = 0⇒ 6x ( x + 1 ) -1 ( x + 1 ) = 0⇒ ( 6x - 1 ) ( x + 1 ) = 0⇒ 6x - 1 = 0 or x + 1 = 0⇒ x = 16 or x = -1Here ( -3 ) x ( -2 ) ≮ 1 ∵ ( -3 ) x ( -2 ) = 6 > 1
Therefore, x= -1 is not the solution.
When substituting x = 16 in 3x × 2x, we have,
3 x 16 x 2 x 16 = 12 x 13 = 16 < 1.
Hence x = 16 is the solution of the given equation.
What is the principal value of cos-1 -32 ?
Prove the following:
tan-1 x = 12 cos-1 1 - x1 + x , x∈ 0, 1
cos-1 1213 + sin-1 35 = sin-1 5665
Write the value of sin π3 - sin-1 - 12
cot-1 1 + sin x + 1 - sin x 1 + sin x - 1 - sin x = x2, x ∈ 0, π4
Find the value of tan-1 xy - tan-1 x - yx + y
Write the principal value of cos-1 12 - 2 sin-1 - 12 .