Question

what are the values of each of the 6 different trigonometric functions? drag the correct value into each box. trigonometric ratio value sin 60° sin (-300)° cos (π/2) cos (5π/2) tan (-45)° tan 315°

Explanation:

Step1: Calculate $\sin 60^\circ$

Using the unit circle or special triangle, $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

Step2: Calculate $\sin (-300^\circ)$

Use the identity $\sin(-\theta)=-\sin\theta$ and $\sin(360^\circ - \theta)=-\sin\theta$. So $\sin(-300^\circ)=\sin(60^\circ)=\frac{\sqrt{3}}{2}$ (since $-300^\circ + 360^\circ = 60^\circ$).

Step3: Calculate $\cos \frac{\pi}{2}$

From the unit circle, $\cos \frac{\pi}{2} = 0$.

Step4: Calculate $\cos \frac{5\pi}{2}$

$\frac{5\pi}{2}=2\pi+\frac{\pi}{2}$, and $\cos(2\pi + \theta)=\cos\theta$, so $\cos \frac{5\pi}{2}=\cos \frac{\pi}{2}=0$.

Step5: Calculate $\tan (-45^\circ)$

Using $\tan(-\theta)=-\tan\theta$ and $\tan 45^\circ = 1$, so $\tan(-45^\circ)=-1$.

Step6: Calculate $\tan 315^\circ$

$315^\circ = 360^\circ - 45^\circ$, $\tan(360^\circ - \theta)=-\tan\theta$, so $\tan 315^\circ = \tan(-45^\circ)=-1$.

Answer:

• $\sin 60^\circ$: $\boldsymbol{\frac{\sqrt{3}}{2}}$
• $\sin (-300)^\circ$: $\boldsymbol{\frac{\sqrt{3}}{2}}$
• $\cos \frac{\pi}{2}$: $\boldsymbol{0}$
• $\cos \frac{5\pi}{2}$: $\boldsymbol{0}$
• $\tan (-45)^\circ$: $\boldsymbol{-1}$
• $\tan 315^\circ$: $\boldsymbol{-1}$