Write the Pythagorean Identity.

\(\displaystyle sin^2\theta+cos^2\theta=1\)

Reorganize the left side of this equation so that it matches the form:   \(\displaystyle 3sin^2\theta-3\)

Subtract cosine squared theta on both sides.

\(\displaystyle sin^2\theta-1=-cos^2\theta\)

Multiply both sides by 3.

\(\displaystyle 3(sin^2\theta-1=-cos^2\theta)\)

\(\displaystyle 3sin^2\theta-3=-3cos^2\theta\)

Of the six trigonometric functions, four are odd, meaning \(\displaystyle f(-x) = -f(x)\). These four are:

• sin x
• tan x
• cot x
• csc x

That leaves two functions which are even, which means that \(\displaystyle f(-x) = f(x)\). These are:

• cos x
• sec x

Of the aforementioned, only \(\displaystyle sec(-x^2) = -sec(-x^2)\) is incorrect, since secant is an even function, which implies that \(\displaystyle sec(-x^2) = sec(x^2)\)

Rewrite \(\displaystyle sin(-15)\) by odd and even identities.

\(\displaystyle sin(-15)=-sin(15)\)

Use the difference identity of sine, and choose the special angles 45 and 30, since their difference equals to 15.

\(\displaystyle sin (A -B) = ( sin A )( cos B ) - ( cos A )( sin B )\)

\(\displaystyle sin (45 -30) = ( sin 45 )( cos 30) - ( cos45 )( sin 30 )\)

\(\displaystyle sin (45 -30) = (\frac{\sqrt2}{2})(\frac{\sqrt3}{2}) -(\frac{\sqrt2}{2})(\frac{1}{2})\)

\(\displaystyle sin (15) = \frac{\sqrt6}{4} -\frac{\sqrt2}{4}\)

\(\displaystyle -sin(15)=-(\frac{\sqrt6}{4} -\frac{\sqrt2}{4})=\frac{\sqrt2}{4} -\frac{\sqrt6}{4}\)

Write the even and odd identities for sine and cosine.

\(\displaystyle sin(-x)=-sin(x)\)

\(\displaystyle cos(-x)=cos(x)\)

Rewrite the expression \(\displaystyle sin(-30)+cos(-30)\) and evaluate.

\(\displaystyle -sin(30)+cos(30) =-\frac{1}{2}+\frac{\sqrt3}{2}= \frac{\sqrt3-1}{2}\)

In order to simplify \(\displaystyle sec(-45)-sec(-45)\), rewrite the expression after applying the rule of odd-even identities for the secant function.

\(\displaystyle sec(-x)=sec(x)\)

\(\displaystyle sec(-45)-sec(-45)=sec(45)-sec(45)=0\)

Which of the following is equivalent to the expression:

\(\displaystyle cot(-\Theta )\)

Begin by recalling the following identity:

\(\displaystyle cot(-\Theta )=-cot(\Theta)\)

Next, recall the relationship between cotangent and tangent:

\(\displaystyle cot(\Theta)=\frac{1}{tan(\Theta)}\)

As well as the relationship between tangent, sine and cosine

\(\displaystyle Tan(\Theta)=\frac{sin{\Theta}}{cos(\Theta)}\)

So to put it all together, we can pull out the negative sign from our original expression:

\(\displaystyle cot(-\Theta )=-cot(\Theta)\)

Next, we can rewrite our cotangent as tangent

\(\displaystyle -cot(\Theta)=\frac{-1}{tan(\Theta)}\)

Finally, we can change our tangent to sine and cosine, but because we are dealing with the reciprocal of tangent, we will need the reciprocal of our identity.

\(\displaystyle -cot(\Theta)=\frac{-1}{tan(\Theta)}=\frac{-cos(\Theta)}{sin(\Theta)}\)

Making our answer:

\(\displaystyle \frac{-cos(\Theta)}{sin(\Theta)}\)

Beware trap answer:

\(\displaystyle \frac{cos(-\Theta)}{sin(\Theta)}\)

This may look good on the surface, but recall

\(\displaystyle cos(-\Theta)\neq -cos(\Theta),cos(-\Theta)=cos(\Theta)\)

Write the reciprocal identity for cosecant.

\(\displaystyle csc(\theta)= \frac{1}{sin(\theta)}\)

Rewrite the expression and use the double angle identities for sine to simplify.

\(\displaystyle \\\frac{2}{csc(2t)}\\ \\= \frac{2}{\frac{1}{sin(2\theta)}}\\ \\= 2sin(2\theta) \\ \\= 2(2sin(\theta)cos(\theta))\)

\(\displaystyle =4cos(\theta)sin(\theta)\)

Rewirte \(\displaystyle 2sec(10x)\) using the reciprocal identity of cosine.

\(\displaystyle sec(\theta)=\frac{1}{cos(\theta)}\)

\(\displaystyle \\2sec(10x)\\ \\=2\left(\frac{1}{cos(10x)}\right)\\ \\= \frac{2}{cos(10x)}\)

Write the reciprocal/ratio identity for cosecant.

\(\displaystyle csc(\theta)=\frac{1}{sin(\theta)}\)

Replace cosecant with sine.

\(\displaystyle \frac{3}{csc(\theta)}=\frac{3}{\frac{1}{sin(\theta)}}= 3sin(\theta)\)

Rewrite \(\displaystyle \frac{sec(\theta)}{tan(\theta)}-\frac{tan(\theta)}{sec(\theta)}\) in terms of sine and cosine.

\(\displaystyle \frac{sec(\theta)}{tan(\theta)}-\frac{tan(\theta)}{sec(\theta)}= \frac{1}{cos(\theta)}\div\frac{sin(\theta)}{cos(\theta)}-\frac{sin(\theta)}{cos(\theta)}\div \frac{1}{cos(\theta)}\)

Dividing fractions is the same as multiplying the numerator by the reciprocal of the denominator.

\(\displaystyle \frac{1}{cos(\theta)}\times\frac{cos(\theta)}{sin(\theta)}-\frac{sin(\theta)}{cos(\theta)}\times cos(\theta)=\frac{1}{sin(\theta)}-sin(\theta)\)

Multiply the second term by sine to get a common denominator. Then after subtracting the second term from the first you can see that a Pythagorean Identity is in the numerator.

Reducing further we arrive at the final answer.

\(\displaystyle \frac{1}{sin(\theta)}-sin(\theta)=\frac{1-sin^2(\theta)}{sin(\theta)}=\frac{cos^2(\theta)}{sin(\theta)}\)