Given \$"sin" \theta = \sqrt(3)/2\$ for \$90^\circ <\theta <180^\circ\$ find the value of \$"cos" \theta\$ using the pythagorean identity.

The given values are

\$"sin" \theta = \sqrt(3)/2\$ for \$90^\circ <\theta <180^\circ\$

The Pythagorean identity that relates the sin and cos ratio is given by

\$"sin"^2(\theta) + "cos"^2(\theta) = 1\$

Now plug the values in the formula

\$\sqrt(3)/(2) + "cos"^2(\theta) = 1\$

Evaluate the square using the fact that \$("A"/"B")^2 = (A""^2)/("B"^2)\$ then make it simpler

⇒ \$3/4 + "cos"^2(\theta) = 1\$
⇒ \$"cos"^2(\theta) = 1 - (3/4)\$
⇒ \$"cos"^2(\theta) = 1/4\$

Note that since \$\theta\$ is in the 2nd quadrant, where \$"cos"(\theta)\$ is negative, we need to take the negative root.

After simplification

⇒ \$"cos"(\theta) = -\sqrt(1/4)\$
⇒ \$"cos"(\theta) = -1/2\$

Therefore the \$"cos"(\theta) = - 1/2\$.

This is incorrect because it’s the positive value of cos⁡θ, and in the given interval, cosine is negative

\$1/2\$

Incorrect due to invalid cosine values within this particular context

-2

This is correct. It correctly did the calculation by using the formula

\$-1/2\$

Invalid values for cosine render this statement incorrect in the context

2

Option

C

Can you summarize what you’ve understood in the above steps?