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Angle x is such that sin x = a + b and cos x = a − b, where a and b are constants.
(i) Show that a^2 + b^2 has a constant value for all values of x.
(ii) In the case where tan x = 2, express a in terms of b.

Respuesta :

Answer:

Step-by-step explanation:

sin x=a+b

cos x=a-b

sin²x+cos²x=(a+b)²+(a-b)²=2a²+2b²

or 2a²+2b²=1

a²+b²=1/2 which is constant for all values of x

(ii)

[tex]\frac{\sin x}{\cos x}=\frac{a+b}{a-b}\\\tan x=\frac{a+b}{a-b}=2\\a+b=2a-2b\\b+2b=2a-a\\or a=3b[/tex]

yoodyannapolis

[tex]a^2 + b^2[/tex] is constant.

The relation is a=3b when tanx=2

We have equations

(i)sin x=a+b...(1)

cos x=a-b...(2)

Now, squaring both the equations

[tex]sin^2 x=(a+b)^2\\cos^2 x=(a-b)^2\\[/tex]

Adding both the equations

[tex]sin^2x+cos^2x=(a+b)^2+(a-b)^2=2a^2+2b^2\\2a^2+2b^2=1\\a^2+b^2=\frac{1}{2}[/tex]

which is a constant.

Hence [tex]a^2 + b^2[/tex] is a constant.

(ii) sin x=a + b...(1)

cos x=a-b...(2)

Dividing both the equations we get

[tex]tanx=\frac{a+b}{a-b}\\2=\frac{a+b}{a-b}\\2(a-b)=a+b\\2a-a=b+2b\\a=3b\\[/tex]

Therefore a=3b when tanx=2

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