room $\cos^2 \theta$ and $\cos \theta^2$ the same?I mean be the $\sin,\cos, \tan ,\cot ,\sec,\csc$. Are they same? Please aid a brickandmortarphilly.coms noob here.

You are watching: (cos θ + cos θ)2 + (cos θ + cos θ)2


*

*

No, they space not the same.

When you form $\cos^2 \theta$ you actually mean $(\cos \theta)^2$.

When you type $\cos \theta^2$ you median $\cos(\theta ^2)$.


*

The an initial notation is supplied to mean$$\cos^2 \theta = \left( \cos \theta \right)^2$$Your second notation will normally be check out as$$\cos \theta^2 = \cos \left( \theta^2 \right)$$although it is sometimes desired to use the notation in the right-hand side to it is in clear.

They room not the exact same since$$\left( \cos \theta \right)^2 = \cos\theta\cos\theta \ne \cos(\theta\theta) = \cos(\theta^2)$$


*

It"s a matter of syntax.

In handwriting or literature: $\sin^2 x \equiv (\sin x)^2$ and $\cos x^2 \equiv \cos (x^2)$;

however for computer system software, say brickandmortarphilly.comematica:

Cos$^2$ refers to $(\cos x)^2$ when Cos describes $\cos (x^2)$.

Should be an extremely careful.


*

Thanks because that contributing an answer to brickandmortarphilly.comematics stack Exchange!

Please be certain to answer the question. Carry out details and share her research!

But avoid

Asking for help, clarification, or responding to other answers.Making statements based on opinion; earlier them up with referrals or an individual experience.

Use brickandmortarphilly.comJax to style equations. Brickandmortarphilly.comJax reference.

See more: In A Physical Bus Topology, Both Ends Of The Medium Must Be Terminated In Order To Prevent What?

To find out more, see our advice on writing great answers.


write-up Your price Discard

By clicking “Post your Answer”, you agree to our regards to service, privacy policy and cookie policy


Not the price you're spring for? Browse various other questions tagged trigonometry or questioning your very own question.


Trigonometric identity: $\frac \tan\theta1-\cot\theta+\frac \cot\theta1-\tan\theta =1+\sec\theta\cdot\csc\theta$