Question

$if \: x {sin}^{3} \alpha + y {cos}^{3} \alpha = \sin( \alpha ) \cos( \alpha ) \: and \: x \sin( \alpha ) - y \cos( \alpha ) = 0 \: ( \alpha \: not \: is \: equal \: to \: 0)\\ then \: prove \: that \: {x}^{2} + {y}^{2} = 1$
PLEASE SHOW ME FULL MATH OF THIS PROBLEM.

Answer 1

Express a sin θ ± b cos θ in the form

R sin(θ ± α),

where a, b, R and α are positive constants.

Solution:

First we take the "plus" case, (θ + α), to make things easy.

Let

a sin θ + b cos θ ≡ R sin (θ + α)

(The symbol " ≡ " means: "is identically equal to")

Using the compound angle formula from before (Sine of the sum of angles),

sin(A + B) = sin A cos B + cos A sin B,

we can expand R sin (θ + α) as follows:

R sin (θ + α)

≡ R (sin θ cos α + cos θ sin α)

≡ R sin θ cos α + R cos θ sin α

So

a sin θ + b cos θ ≡ R cos α sin θ + R sin α cos θ

Equating the coefficients of sin θ and cos θ in this identity, we have:

For sin θ:

a = R cos α ..........(1)
(in green above)

For cos θ:

b = R sin α .........(2)
(in red above)

Eq. (2) ÷ Eq.(1):

\displaystyle\frac{b}{{a}}=\frac{{{R} \sin{\alpha}}}{{{R} \cos{\alpha}}}= \tan{\alpha}
a
b
​
=
Rcosα
Rsinα
​
=tanα
So

\displaystyle\alpha= \arctan{\ }\frac{b}{{a}}α=arctan
a
b
​

(α is a positive acute angle and a and b are positive.)

Now we square each of Eq. (1) and Eq. (2) and add them to find an expression for R.

[Eq. (1)]2 + [Eq. (2)]2:

a2 + b2

= R2 cos2 α + R2 sin2 α

= R2(cos2 α + sin2 α)

= R2

(since cos2 A + sin2 A = 1)

So

\displaystyle{R}=\sqrt{{{a}^{2}+{b}^{2}}}R=
a
2
+b
2

​

(We take only the positive root)

In summary, if

\displaystyle\alpha= \arctan{\ }\frac{b}{{a}}α=arctan
a
b
​

and

\displaystyle{R}=\sqrt{{{a}^{2}+{b}^{2}}}R=
a
2
+b
2

​

then we have expressed a sin θ + b cos θ in the form required:

a sin θ + b cos θ = R sin(θ + α)

Answer 2

HERE is ur answer meta✌️✌️



Express a sin θ ± b cos θ in the form

R sin(θ ± α),

where a, b, R and α are positive constants.

Solution:

First we take the "plus" case, (θ + α), to make things easy.

Let

a sin θ + b cos θ ≡ R sin (θ + α)

(The symbol " ≡ " means: "is identically equal to")

Using the compound angle formula from before (Sine of the sum of angles),

sin(A + B) = sin A cos B + cos A sin B,

we can expand R sin (θ + α) as follows:

R sin (θ + α)

≡ R (sin θ cos α + cos θ sin α)

≡ R sin θ cos α + R cos θ sin α

So

a sin θ + b cos θ ≡ R cos α sin θ + R sin α cos θ

Equating the coefficients of sin θ and cos θ in this identity, we have:

For sin θ:

a = R cos α ..........(1)
(in green above)

For cos θ:

b = R sin α .........(2)
(in red above)

Eq. (2) ÷ Eq.(1):

\displaystyle\frac{b}{{a}}=\frac{{{R} \sin{\alpha}}}{{{R} \cos{\alpha}}}= \tan{\alpha}
a
b
​
=
Rcosα
Rsinα
​
=tanα
So

\displaystyle\alpha= \arctan{\ }\frac{b}{{a}}α=arctan
a
b
​

(α is a positive acute angle and a and b are positive.)

Now we square each of Eq. (1) and Eq. (2) and add them to find an expression for R.

[Eq. (1)]2 + [Eq. (2)]2:

a2 + b2

= R2 cos2 α + R2 sin2 α

= R2(cos2 α + sin2 α)

= R2

(since cos2 A + sin2 A = 1)

So

\displaystyle{R}=\sqrt{{{a}^{2}+{b}^{2}}}R=
a
2
+b
2

​

(We take only the positive root)

In summary, if

\displaystyle\alpha= \arctan{\ }\frac{b}{{a}}α=arctan
a
b
​

and

\displaystyle{R}=\sqrt{{{a}^{2}+{b}^{2}}}R=
a
2
+b
2

​

then we have expressed a sin θ + b cos θ in the form required:

a sin θ + b cos θ = R sin(θ + α)


HOPE IT HELPS U ✌️✌️☺️