Is (8sin^2(t))/(tan^2(t))=8cos^2(t) ?

The answer to whether (8sin^2(t))/(tan^2(t))=8cos^2(t) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (8sin^2(t))/(tan^2(t))=8cos^2(t)

prove

\frac{8 sin ^{2} ( t )}{tan ^{2} ( t )} = 8 cos^{2} (t)

True

Solution steps

\frac{8 sin ^{2} ( t )}{tan ^{2} ( t )} = 8 cos^{2} (t)

Manipulating left side

\frac{8 sin ^{2} ( t )}{tan ^{2} ( t )}

Express with sin, cos

\frac{8 sin ^{2} ( t )}{tan ^{2} ( t )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{8 sin ^{2} ( t )}{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}}

Simplify

\frac{8 sin ^{2} ( t )}{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}} : 8 cos^{2} (t)

\frac{8 sin ^{2} ( t )}{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}}

(\frac{sin ( t )}{cos ( t )})^{2} = \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

(\frac{sin ( t )}{cos ( t )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( t )}{cos ^{2} ( t )}

= \frac{8 sin ^{2} ( t )}{\frac{s i n ^{2} ( t )}{c o s ^{2} ( t )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{8 sin ^{2} ( t ) cos ^{2} ( t )}{sin ^{2} ( t )}

Cancel the common factor:

sin^{2} (t)

= 8 cos^{2} (t)

= 8 cos^{2} (t)

= 8 cos^{2} (t)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( θ ) + sin ( 3 θ )}{2 cos ( θ )} = sin (2 θ)

cos (4 θ) - 4 cos (2 θ) = 8 sin^{4} (θ) - 3

tan^{2} (θ) = \frac{1}{cos ^{2} ( θ )} - 1

sin (x) cos (x) tan (x) = 1 - cos^{2} (x)

\frac{1 + tan ( u )}{1 + cot ( u )} = tan (u)

Is (8sin^2(t))/(tan^2(t))=8cos^2(t) ?
The answer to whether (8sin^2(t))/(tan^2(t))=8cos^2(t) is True