Question

I’m exercise 5, write the expression in terms of sine and/or cosine Number 5 I’m a little unsure about. I circled number 5 for you to see.

Answer 1

`\begin{gathered} \tan \text{ 60}\degree\text{ = }\frac{\sin\text{ 60}\degree}{\cos\text{ 60}\degree} \\ or \\ \tan \text{ 60}\degree\text{ = }\frac{\text{cos(90 - 60)}\degree}{\text{sin(90 - 60)}\degree} \end{gathered}` Explanation:
`\begin{gathered} 5)\text{ Given: tan60}\degree \\ To\text{ write in terms of sine and/or cosine} \end{gathered}`

We need to find the relationship between sine, cosine and tangent

`In\text{ trigonometry, tan }\theta\text{ = }\frac{\sin \theta\text{ }}{\cos \text{ }\theta}`
`\begin{gathered} In\text{ this case }\theta\text{= 60}\degree \\ We\text{ can replace our }\theta\text{ with 60}\degree \\ \tan \text{ 60}\degree\text{ = }\frac{\sin\text{ 60}\degree}{\cos\text{ 60}\degree} \end{gathered}`
`\begin{gathered} \sin \theta\text{ = }cos(90\text{ - }\theta) \\ \sin \text{ 60 = cos(90 - 60)} \\ \\ \cos \theta\text{ = }\sin (90\text{ - }\theta) \\ cos\text{ 60 = sin(90 - 60)} \\ \\ \tan \text{ 60}\degree\text{ = }\frac{\sin\text{ 60}\degree}{\cos\text{ 60}\degree}\text{ = }\frac{\text{cos(90 - 60)}\degree}{\text{sin(90 - 60)}\degree} \end{gathered}`