Advertisements
Advertisements
Question
Prove the following:
3 sin−1 x = sin−1 (3x − 4x3), `x ∈ [-1/2, 1/2]`
Advertisements
Solution
Let x = sin θ.
Then, sin−1 x = θ.
We have,
R.H.S = sin−1 (3x – 4x3) = sin−1 (3 sin θ – 4 sin3θ)
= sin−1 (sin 3θ) = sin−1 (3 sin θ – 4 sin3θ)
= 3θ = sin−1 (3 sin θ – 4 sin3θ)
= 3 sin−1 x = sin−1 (3 sin θ – 4 sin3θ)
R.H.S = L.H.S
APPEARS IN
RELATED QUESTIONS
Solve for x : tan-1 (x - 1) + tan-1x + tan-1 (x + 1) = tan-1 3x
Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`
Write the following function in the simplest form:
`tan^(-1) (sqrt((1 - cos x)/(1 + cos x)))`, 0 < x < π
Write the function in the simplest form: `tan^(-1) ((cos x - sin x)/(cos x + sin x)) `,` 0 < x < pi`
Write the following function in the simplest form:
`tan^(-1) ((3a^2 x - x^3)/(a^3 - 3ax^2)), a > 0; (-a)/sqrt(3) < x < a/sqrt(3)`
if `tan^(-1) (x-1)/(x - 2) + tan^(-1) (x + 1)/(x + 2) = pi/4` then find the value of x.
Prove that `cot^(-1) ((sqrt(1 + sin x) + sqrt(1 - sinx))/(sqrt(1 + sin x) - sqrt(1 - sinx))) = x/2, x ∈ (0, pi/4)`.
Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
If y = `(x sin^-1 x)/sqrt(1 -x^2)`, prove that: `(1 - x^2)dy/dx = x + y/x`
Find: ∫ sin x · log cos x dx
Solve: tan-1 4 x + tan-1 6x `= π/(4)`.
Find the value of the expression in terms of x, with the help of a reference triangle
sin (cos–1(1 – x))
Find the value of the expression in terms of x, with the help of a reference triangle
cos (tan–1 (3x – 1))
Prove that `sin^-1 3/5 - cos^-1 12/13 = sin^-1 16/65`
Simplify: `tan^-1 x/y - tan^-1 (x - y)/(x + y)`
Choose the correct alternative:
`sin^-1 3/5 - cos^-1 13/13 + sec^-1 5/3 - "cosec"^-1 13/12` is equal to
Choose the correct alternative:
`tan^-1 (1/4) + tan^-1 (2/9)` is equal to
Choose the correct alternative:
The equation tan–1x – cot–1x = `tan^-1 (1/sqrt(3))` has
Choose the correct alternative:
If `sin^-1x + cot^-1 (1/2) = pi/2`, then x is equal to
Evaluate: `sin^-1 [cos(sin^-1 sqrt(3)/2)]`
If α ≤ 2 sin–1x + cos–1x ≤ β, then ______.
If `sin^-1 ((2"a")/(1 + "a"^2)) + cos^-1 ((1 - "a"^2)/(1 + "a"^2)) = tan^-1 ((2x)/(1 - x^2))`. where a, x ∈ ] 0, 1, then the value of x is ______.
The value of the expression `tan (1/2 cos^-1 2/sqrt(5))` is ______.
Solve for x : `"sin"^-1 2 "x" + sin^-1 3"x" = pi/3`
`"sin" {2 "cos"^-1 ((-3)/5)}` is equal to ____________.
`"cot" ("cosec"^-1 5/3 + "tan"^-1 2/3) =` ____________.
If `"tan"^-1 2 "x + tan"^-1 3 "x" = pi/4`, then x is ____________.
`"cos"^-1["cos"(2"cot"^-1(sqrt2 - 1))]` = ____________.
`"sin"^-1 (1 - "x") - 2 "sin"^-1 "x" = pi/2`
If `"sin"^-1 (1 - "x") - 2 "sin"^-1 ("x") = pi/2,` then x is equal to ____________.
If `3 "sin"^-1 ((2"x")/(1 + "x"^2)) - 4 "cos"^-1 ((1 - "x"^2)/(1 + "x"^2)) + 2 "tan"^-1 ((2"x")/(1 - "x"^2)) = pi/3` then x is equal to ____________.
The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:
Measure of ∠DAB = ________.
`tan^-1 1/2 + tan^-1 2/11` is equal to
If `cos^-1(2/(3x)) + cos^-1(3/(4x)) = π/2(x > 3/4)`, then x is equal to ______.
`tan^-1 sqrt3 - cot^-1 (- sqrt3)` is equal to ______.
Solve for x: `sin^-1(x/2) + cos^-1x = π/6`
