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3 years ago

Find the sum of 3x + y , - 4x -2y

Mathematics

1 answer:

ser-zykov [4K]3 years ago

3 0

Answer:

(-x-y)

Step-by-step explanation:

Given data

First expression = ( 3x + y)

Second expression = (-4x -2y)

Add the two expresssion

= ( 3x + y)+ (-4x -2y)

Open bracket

=3x+y-4x-2y

collect like terms

=3x-4x+y-2y

=-x-y

Hence the result is (-x-y)

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3 years ago

Use the power reduction formulas to rewrite the expression. (Hint: Your answer should not contain any exponents greater than 1.)

Lapatulllka [165]

Some useful relations and identities:

$\tan x=\dfrac{\sin x}{\cos x}$

$\sin^2x=\dfrac{1-\cos2x}2$

$\cos^2x=\dfrac{1+\cos2x}2$

By the first relation, we have

$\tan^2x\sin^3x=\dfrac{\sin^5x}{\cos^2x}=\dfrac{(\sin^2x)^2\sin x}{\cos^2x}$

Applying the two latter identities, we get

$\dfrac{\left(\frac{1-\cos2x}2\right)^2\sin x}{\frac{1+\cos2x}2}=\dfrac{\frac{1-2\cos2x+\cos^22x}4\sin x}{\frac{1+\cos2x}2}=\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}$

We can apply the third identity again:

$\dfrac{(1-2\cos2x+\cos^22x)\sin x}{2(1+\cos2x)}=\dfrac{\left(1-2\cos2x+\frac{1+\cos4x}2\right)\sin x}{2(1+\cos2x)}=\dfrac{(3-4\cos2x+\cos4x)\sin x}{4(1+\cos2x)}$

and this is probably as far as you have to go, but by no means is it the only possible solution.

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