How to Solve Trigonometric Functions When \( \cos(t) = -\frac{9}{10} \)? (2024)

Hello,

I am trying to solve this. This material is not covered in my class, but I still want to know how to do it.
If cos(t)=$\frac{-9}{10}$ where $\pi$ <t<$\frac{3\pi}{2}$ find the values of

cos(2t)=
sin(2t)=
cos($\frac{t}{2}$)=
sin($\frac{t}{2}$)=

Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression.

I have no idea how to even start with it. I will appreciate your help!

lastochka said:

Hello,

I am trying to solve this. This material is not covered in my class, but I still want to know how to do it.
If cos(t)=$\frac{-9}{10}$ where $\pi$ <t<$\frac{3\pi}{2}$ find the values of

cos(2t)=
sin(2t)=
cos($\frac{t}{2}$)=
sin($\frac{t}{2}$)=

Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression.

I have no idea how to even start with it. I will appreciate your help!

See Also

4.3: The Trigonometric Functions - Unit Circle DefinitionAnatomy Drawing LessonsFind The Values Of The Trigonometric Ratios For X Using The Graph Of The Unit Circle.In The Text It Is Asserted That For Any (b1,b2,b3) The Vectors (1,1,1),(0,1,1),(0,0,1), And (b1,b2,b3)

You are expected to use the following identities:

$\displaystyle \begin{align*} \sin^2{(x)} + \cos^2{(x)} &\equiv 1 \\ \\ \cos{(2x)} &\equiv \cos^2{(x)} - \sin^2{(x)} \\ &\equiv 2\cos^2{(x)} - 1 \\ &\equiv 1 - 2\sin^2{(x)} \\ \\ \sin{(2x)} &\equiv 2\sin{(x)}\cos{(x)} \end{align*}$

See what you can do...

Prove It said:

You are expected to use the following identities:

$\displaystyle \begin{align*} \sin^2{(x)} + \cos^2{(x)} &\equiv 1 \\ \\ \cos{(2x)} &\equiv \cos^2{(x)} - \sin^2{(x)} \\ &\equiv 2\cos^2{(x)} - 1 \\ &\equiv 1 - 2\sin^2{(x)} \\ \\ \sin{(2x)} &\equiv 2\sin{(x)}\cos{(x)} \end{align*}$

See what you can do...

Well, I know these identities, but as I said I am not sure how to use them in this question. Can you elaborate? Thanks

I finally did this problem and the answers are correct. Posting just in case anyone else is interested

cos(2t)=$\frac{31}{50}$

sin(2t)=$\frac{9\sqrt{19}}{50}$

cos(t2)=$\frac{-\sqrt{5}}{10}$

sin(t2)=$\frac{\sqrt{95}}{10}$

This is how I would work the problem...we know $t$ is in quadrant III, and so sine and cosine of $t$ are both negative.

\(\displaystyle \sin(t)=-\sqrt{1-\left(-\frac{9}{10}\right)^2}=-\sqrt{\frac{19}{100}}=-\frac{\sqrt{19}}{10}\)

Now, we find:

\(\displaystyle \cos(2t)=\frac{81}{100}-\frac{19}{100}=\frac{62}{100}=\frac{31}{50}\)

\(\displaystyle \sin(2t)=2\left(-\frac{\sqrt{19}}{10}\right)\left(-\frac{9}{10}\right)=\frac{9\sqrt{19}}{50}\)

\(\displaystyle \cos\left(\frac{t}{2}\right)=-\sqrt{\frac{1-\frac{9}{10}}{2}}=-\frac{1}{2\sqrt{5}}\)

\(\displaystyle \sin\left(\frac{t}{2}\right)=\sqrt{\frac{1+\frac{9}{10}}{2}}=\frac{\sqrt{19}}{2\sqrt{5}}\)