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2 years ago

14

How do I do this I need help right now

1 answer:

kodGreya [7K]2 years ago

6 0

To answer this (specifically the first one) you have to count how many spaces each point moves. (How many spaces up/down does it go? Left/right?) then you put rise (up/down) over run (left/right) and you should get the slope!

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2 years ago

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago