Question

how to write square n using keyboard

Answer 1

n^2

Answer 2

or you can use latex on here like \( 3^2 \)

Answer 3

Yeah, on here you can use that power, but in general it's n^2

Answer 4

type \ ( 3^2 \ ) with no spaces

Answer 5

You can even \( 4^\frac{1}{2} \) = \ ( 4^\frac{1}{2} \ ) \ ( and \ ) cannot have spaces

Answer 6

Alternatively, if you're using a Mac you can open the character palette and select from a bunch of mathematical symbols. If I remember correctly, you can open it with Ctrl+Alt+T, but I could be wrong.

If you're using a PC, and your keyboard has a numpad on the right side, holding Alt and typing 01478 will give you an exponent 2, 01479 gives you a 3, and 252 gives you an 'n'

Answer 7

\[\sin5A= 5\cos^4A timessinA -10\cos^2A timessin^3A+\sin^5A \]

Answer 8

\[\sin5a=5\cos^4a\sin a-10\cos^2a\sin^3a+\sin^5a\]
Do you have to prove the identity?

Answer 9

yes

Answer 10

plzz rply me soon

Answer 11

Alright, so I think working with the left side would be best. It's tedious, but you should be able to decompose the \(5a\) into \(3a\) and \(2a\), and eventually \(a\) using the angle sum identities:
\[\sin(x+y)=\sin x\cos y+\sin y\cos x\\
\cos(x+y)=\cos x\cos y-\sin x\sin y\]
\[\begin{align*}LHS&=\sin5a\\
&=\sin(3+2)a\\
&=\sin3a\cos2a+\sin2a\cos3a\\
&=\sin(2+1)a\cos2a+\sin2a\cos(2+1)a\\
&=(\sin2a\cos a+\sin a\cos2a)\cos2a+(\cos2a\cos a-\sin 2a\sin a)\sin 2a\end{align*}\]
Then you expand using the double angle identities.
\[\sin2x=2\sin x\cos x\\
\cos2x=\cos^2x-\sin^2x\]

Answer 12

You should get
\[(2\sin a\cos^2a+\cos^2a\sin a-\sin^3a)(\cos^2a-\sin^2a)+\\
~~~~~~~(\cos^3a-\sin^2a\cos a-2\sin^2a\cos a)(2\sin a \cos a)\]
To clean it up, I'll use a substitution, \(x=\sin a\) and \(y=\cos a\):
\[(2xy^2+xy^2-x^3)(y^2-x^2)+(y^3-x^2y-2x^2y)2xy\]
Some terms disappear:
\[(3xy^2-x^3)(y^2-x^2)+(y^3-3x^2y)2xy\]
Now you distribute:
\[\left(3xy^4-x^3y^2-3x^3y^2+x^5\right)+\left(2xy^4-6x^3y^2\right)\]
Simplify:
\[5xy^4-10x^3y^2+x^5\]
Undo the substitution:
\[LHS=5\sin a\cos^4a-10\sin^3a\cos^2a+\sin^5a\]