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3 arccos(x) = arccos(4x3 - 3x)
We will learn how to prove the property of the inverse trigonometric function 3 arccos(x) = arccos(4x3 - 3x) or, 3 cos−1 x = cos−1 (4x3 - 3x)
Proof:
Let, cos−1 x = θ
Therefore, cos θ = x
Now we know that, sin 3θ = 4 cos3 θ - 3 cos θ
⇒ cos 3θ = 4x3 - 3x
Therefore, 3θ = cos−1 (4x3 - 3x)
⇒ 3 cos−1 x = cos−1 (4x3 - 3x)
or, 3 arccos(x) = arccos(4x3 - 3x). Proved.
● Inverse Trigonometric Functions
- General and Principal Values of sin−1 x
- General and Principal Values of cos−1 x
- General and Principal Values of tan−1 x
- General and Principal Values of csc−1 x
- General and Principal Values of sec−1 x
- General and Principal Values of cot−1 x
- Principal Values of Inverse Trigonometric Functions
- General Values of Inverse Trigonometric Functions
- arcsin(x) + arccos(x) = \frac{π}{2}
- arctan(x) + arccot(x) = \frac{π}{2}
- arctan(x) + arctan(y) = arctan(\frac{x + y}{1 - xy})
- arctan(x) - arctan(y) = arctan(\frac{x - y}{1 + xy})
- arctan(x) + arctan(y) + arctan(z)= arctan\frac{x + y + z – xyz}{1 – xy – yz – zx}
- arccot(x) + arccot(y) = arccot(\frac{xy - 1}{y + x})
- arccot(x) - arccot(y) = arccot(\frac{xy + 1}{y - x})
- arcsin(x) + arcsin(y) = arcsin(x \sqrt{1 - y^{2}} + y\sqrt{1 - x^{2}})
- arcsin (x) - arcsin(y) = arcsin (x \sqrt{1 - y^{2}} - y\sqrt{1 - x^{2}})
- arccos (x) + arccos(y) = arccos(xy - \sqrt{1 - x^{2}}\sqrt{1 - y^{2}})
- arccos(x) - arccos(y) = arccos(xy + \sqrt{1 - x^{2}}\sqrt{1 - y^{2}})
- 2 arcsin(x) = arcsin(2x\sqrt{1 - x^{2}})
- 2 arccos(x) = arccos(2x^{2} - 1)
- 2 arctan(x) = arctan(\frac{2x}{1 - x^{2}}) = arcsin(\frac{2x}{1 + x^{2}}) = arccos(\frac{1 - x^{2}}{1 + x^{2}})
- 3 arcsin(x) = arcsin(3x - 4x^{3})
- 3 arccos(x) = arccos(4x^{3} - 3x)
- 3 arctan(x) = arctan(\frac{3x - x^{3}}{1 - 3 x^{2}})
- Inverse Trigonometric Function Formula
- Principal Values of Inverse Trigonometric Functions
- Problems on Inverse Trigonometric Function
11 and 12 Grade Math
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