Given one trigonometric function, find another

Given one trig ratio and the quadrant, draw a triangle, use Pythagoras for the third side, and use ASTC to get the sign of every other ratio.

The method

Example 1●●

Given $\sin\theta = \dfrac{3}{5}$ and $90^\circ < \theta < 180^\circ$, find $\cos\theta$ and $\tan\theta$.

Third side: $\sqrt{5^2 - 3^2} = 4$.

Opposite 3, hypotenuse 5.

$\cos\theta = -\dfrac{4}{5}$ and $\tan\theta = -\dfrac{3}{4}$.

Quadrant 2: only sin is positive.

Example 2●●●

Given $\tan\theta = -2$, find the possible values of $\cos\theta$.

$\theta$ is in quadrant 2 or quadrant 4.

tan is negative in quadrants 2 and 4.

Hypotenuse $\sqrt{2^2 + 1^2} = \sqrt{5}$, so the size of $\cos\theta$ is $\dfrac{1}{\sqrt{5}}$.

Opposite 2, adjacent 1.

Quadrant 2: $\cos\theta = -\dfrac{1}{\sqrt{5}}$. Quadrant 4: $\cos\theta = \dfrac{1}{\sqrt{5}}$.

One point gives all three ratios

Drag P. Its coordinates and its distance r from the origin give sin θ = y/r, cos θ = x/r and tan θ = y/x.

Try this

Put $P$ at $(-4, 3)$. Compare the readout with Example 1.

Flashcards

Step 1 when given one ratio

Draw the related-angle triangle, ignoring signs.

Step 2

Pythagoras for the third side.

Step 3

Sign from the quadrant (ASTC).

Practice