Question

Evaluate : sin 45°- cos 36°

Answer 1

Answer:

$\Large\boxed{\frac{\sqrt{2} }{2}-\frac{\sqrt{5+1} }{4}=-0.10}$

EXPLANATIONS:

Question:  ⇒ Evaluate: sin 45°- cos 36°

Given:

Sin (45°)- cos (36°)

Solutions:

First, trivial identify by sin.

$\displaystyle \sin(45)^{\circ}=\frac{\sqrt{2} }{2}$

$\displaystyle \cos(36)^{\circ}-\sin(18)^{\circ}=\frac{1}{2}$

$\Large\boxed{\textnormal{PRODUCT TO SUM IDENTIFY}}$

$\Large\boxed{2\sin(X)\cos(Y)=\sin(X+Y)-\sin(X-Y)}$

$\displaystyle 2\cos(36)^{\circ}\sin(18)^{\circ}=\sin(54)^{\circ}-\sin(18)^{\circ}$

$\Large\boxed{\textnormal{DOUBLE ANGLE IDENTIFY}}$

$\displaystyle \sin(2x)=2\sin(x)\cos(x)$

$\displaystyle \sin \left(72^{\circ \:}\right)=2\sin \left(36^{\circ \:}\right)\cos \left(36^{\circ \:}\right)$

$\displaystyle \sin \left(36^{\circ \:}\right)=2\sin \left(18^{\circ \:}\right)\cos \left(18^{\circ \:}\right)$

Secondly, you have to multiply by 2 equations from left to right.

$\displaystyle =\sin \left(72^{\circ \:}\right)\sin \left(36^{\circ \:}\right)=4\sin \left(36^{\circ \:}\right)\sin \left(18^{\circ \:}\right)\cos \left(36^{\circ \:}\right)\cos \left(18^{\circ \:}\right)$

Thirdly, divide by sin (36)° from both sides.

$\displaystyle \sin(72)^{\circ}=4\sin(18)^{\circ}\cos(36)^{\circ}\cos(18)^{\circ}$

$\displaystyle \sin(x)=\cos(90^{\circ}-x)$

Solve.

$\displaystyle \sin(72)=\cos(90-72)$

$\displaystyle \cos(90-72)=4\sin(18)\cos(36)\cos(18)$

Subtract.

$\displaystyle 90-72=18$

$\displaystyle \cos(18)=4\sin(18)\cos(36)\cos(18)$

Next, divide by cos(18) from both sides.

$\displaystyle 1=4\sin(18)\cos(36)$

Divide by 2 from both sides.

$\displaystyle \frac{1}{2}=2\sin(18)\cos(36)$

Solve with 1/2=2sin(18)cos(36).

$\displaystyle \frac{1}{2}=\sin(54)-\sin(18)$

$\displaystyle \sin(54)=\cos(90-54)$

$\displaystyle \frac{1}{2}=\cos \left(90^{\circ \:}-54^{\circ \:}\right)-\sin \left(18^{\circ \:}\right)$

Subtract.

$\displaystyle 90-54=36$

$\displaystyle \frac{1}{2}=\cos(36)-\sin(18)$

Add equations from left to right.

$\displaystyle \cos(36)+\sin(18)=\frac{\sqrt{5} }{2}$

$\displaystyle \cos(36)-\sin(18)=\frac{1}{2}$

$\displaystyle (\cos (36^{\circ \:}\right)+\sin \left(18^{\circ \:}\right)\right)+\left(\cos \left(36^{\circ \:}\right)-\sin \left(18^{\circ \:}\right)\right)\right)=\left(\frac{\sqrt{5}}{2}+\frac{1}{2}\right)$

Solve. (Refine and simplify.)

$\displaystyle \cos(36)=\frac{\sqrt{5+1} }{4}$

$\displaystyle \cos(36)=\frac{\sqrt{5+1} }{4}=\boxed{{\frac{\sqrt{2} }{2}-\frac{\sqrt{5+1} }{4}=-0.10}}$

$\Large\boxed{\frac{\sqrt{2} }{2}-\frac{\sqrt{5+1} }{4}=-0.10}$

Therefore, the correct answer is √2/2-√5+1/4=-0.10.