# Inverse trigonometric functions without calculator, arcsin, arctan

How can I evaluate this expression without the use of a calculator and only assuming i know the standard angles ($$\pi/3, \pi/4, \pi/6$$)?

\begin{align*} \arcsin\left(\frac{5\sqrt{3}}{14}\right)-\arctan\left(\frac{1}{4\sqrt{3}}\right) \end{align*}

Let $$\arcsin\dfrac{5\sqrt3}{14}=y,0

$$\sin y=?$$

$$\cos y=+\sqrt{1-\sin^2y}=?$$

$$\tan y=?$$

$$y=\arctan?$$

Now $$-\arctan(a)=\arctan(-a)$$

Hint: Draw a right triangle $$\triangle ABC$$ and label the lengths of two sides appropriately so that the base angle $$\alpha=\arcsin\left(\frac{5\sqrt{3}}{14}\right)$$ as in the diagram below. Then, by the Pythagorean Theorem, the remaining side must have length $$11$$.

Follow the example of the first triangle to appropriately label two sides of the second triangle $$\triangle DEF$$ so that the base angle at vertex $$D$$ is $$\delta=\arctan\left(\frac{1}{4\sqrt{3}}\right)$$. Use the Pythagorean Theorem to find the length of the third side.

Now the task is reduced to finding the angle $$\alpha-\delta$$. We know that both angles are positive acute angles and that $$\alpha$$ is the larger of the two, so we know that $$\alpha-\delta$$ is a positive acute angle and is one of the special angles which were mentioned in the problem.

Suggestion: Find $$\sin(\alpha-\delta)$$ using an identity which you studied and use the two triangles to find the values needed in the formula. You should find that $$\sin(\alpha-\delta)$$ is the sine of one of the special angles. Then you will have your answer.