TRIGONOMETRY · HL
Inverse Trig
Cos⁻¹ is really an angle.
Section 1 of 5
Find Sin 30°
$\sin 30^\circ = \dfrac{1}{2} = 0.5$
Section 2 of 5
Find A which Cos A = ½
$\cos^{-1} \dfrac{1}{2} = A$
$60 = A \qquad \cos 60 = \dfrac{1}{2}$
$\cos^{-1} x = A \quad \Rightarrow \quad$ cos inverse of $x$.
$\cos^{-1}$ is really an angle.
Section 3 of 5
Operation · Inverse
OperationInverse
AddSubtract
MultiplyDivide
IndicesLogs
Cos$\cos^{-1}$
Section 4 of 5
Find Sin(Cos⁻¹ x)
$\cos A = x \qquad \cos^{-1} x = A$
$\cos^{-1} x = A \qquad$ really need $\operatorname{Sin}A$
$\cos A = \dfrac{x}{1} = \dfrac{\text{adj}}{\text{hyp}}$
$\operatorname{Sin}A = \sqrt{1 - x^{2}}$
Section 5 of 5
Find Tan(Sin⁻¹ x)
Let $\quad \operatorname{Sin}^{-1} x = A$
$\operatorname{Sin}A = \dfrac{x}{1} = \dfrac{\text{opp}}{\text{hyp}}$
Really need
$\operatorname{Tan}A = \dfrac{x}{\sqrt{1 - x^{2}}}$
SUM
The lot in one box
Inverse trig recap
1.$\cos^{-1} x = A$ — $\cos^{-1}$ is really an angle.
2.Operation · Inverse: Add·Subtract, Multiply·Divide, Indices·Logs, Cos·$\cos^{-1}$.
3.$\operatorname{Sin}(\cos^{-1} x) = \sqrt{1 - x^{2}}$
4.$\operatorname{Tan}(\operatorname{Sin}^{-1} x) = \dfrac{x}{\sqrt{1 - x^{2}}}$
End of lesson
Inverse Trig — HL · Mathslive.ie