Question

If a sin cube x+ b cos cube x=sin x cos x and a sin x = b cos x, then find a square + b square​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• $\sf a sin^3 x+bcos^3x=sinxcosx$

• $\sf asinx=bcosx$

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• $\sf a^2+b^2=?$

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

PART - 1.

$\to \sf a sin^3 x+bcos^3x=sinxcosx$

▣ This can be written as:

$\to \sf sin^2(a sin x)+bcos^3x=sinxcosx$

▣ Substitute $\sf asinx=bcosx$

$\to \sf sin^2(bcosx)+bcos^3x=sinxcosx$

▣ Take b cos x common.

$\to \sf bcosx(sin^2+cos^2)=sinxcosx$

▣ We know that $\sf sin^2A+cos^2A=1$. Therefore, we get:

$\to \sf bcosx=sinxcosx$

$\to \sf \dfrac{bcosx}{sinxcosx}=1$

▣ Further solving this, and cancelling cosx and transposing sinx to the RHS, we get  

b=sinx.

PART - 2.

$\to \sf a sin^3 x+bcos^3x=sinxcosx$

▣ This can be written as:

$\to \sf a sin^3 x+cos^2(bcosx)=sinxcosx$

▣ Substitute $\sf asinx=bcosx$

$\to \sf a sin^3 x+cos^2(asinx)=sinxcosx$

▣ Take a sin x common.

$\to \sf asinx(sin^2+cos^2)=sinxcosx$

▣ We know that $\sf sin^2A+cos^2A=1$. Therefore, we get:

$\to \sf asinx=sinxcosx$

$\to \sf \dfrac{asinx}{sinxcosx}=1$

▣ Further solving this, and cancelling sinx and transposing cosx to the RHS, we get a=cosx.

PART 3.

Now, we know that:

• a=cosx.
• b=sinx.

Then:

$\sf \to a^2+b^2$

Substitute the values.

$\sf \to cos^2x+sin^2x$

$\leadsto \sf{\red{ 1}}$

∴ a²+b²=1.

FUNDAMENTAL TRIGONOMETRIC IDENTITIES:

$\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}$