Question

sin= 4/5. find the sin2 and cos2?

Answer 1

Answer:

$\rm{\sin^{2} = \dfrac{16}{25} \: \: and \: \: \cos^{2} = \dfrac{9}{25}}$

Step-by-step explanation:

Given:

$\rm{\sin = \dfrac{4}{5}}$

To find:

${\sin}^{2}$ and ${\cos}^{2}$

Solution :

• ${\sin}^{2}$

$\longrightarrow \: {\sin}^{2} = \bigg( \dfrac{4}{5}\bigg)^{2}$

$\longrightarrow \: {\sin}^{2} = \dfrac{16}{25}$

___________________

• ${\cos}^{2}$

We know that,

$\rm{ \sin = \dfrac{Opposite \: side}{Hypotenuse}}$

So, In the given ∆ABC opposite side to angle theta is AB and hypothenuse is AC.

• AB = 4 units
• AC = 5 units

By Pythagoras theorem,

$\longrightarrow$ (Hypotenuse)² = (Base)² + (Height)²

$\longrightarrow$ (5)² = (BC)² + (4)²

$\longrightarrow$ 25 = (BC)² + 16

$\longrightarrow$ BC² = 25 - 16

$\longrightarrow$ BC = √9

$\longrightarrow$ BC = 3 units.

$\rm{{\cos} = \dfrac{Adjacent \: Side}{Hypotenuse}}$

• Adjacent side = 3 units
• Hypotenuse = 5 units

$\longrightarrow \: {\cos}^{2} = \bigg( \dfrac{3}{5}\bigg)^{2}$

$\longrightarrow \: {\cos}^{2} = \dfrac{9}{25}$

$\rm{\therefore \: \sin^{2} = \dfrac{16}{25} \: \: and \: \: \cos^{2} = \dfrac{9}{25}}$

Answer 2

$\huge\red{ \boxed{ \boxed{ \bf \blue{{sin}^{2} = \frac{16}{25} }}}} \\ \\ \huge\color{orange}{ \boxed { \boxed{ \bf \green{ {cos}^{2} = \frac{9}{25} }}}}$

Step-by-step explanation:

$\bf \sin = \frac{4}{5} \\ \\ \bf so \\ \bf { \sin}^{2} = { \bigg( \frac{4}{5} \bigg)}^{2} \\ \\ \bf = \frac{16}{25}$

$\bf sin = \frac{4}{5} = \frac{4x}{5x}$

We know that,

In a right triangle,

$\bf sin = \frac{height}{hypotenuse}$

=> height = 4x ; hypotenuse = 5x

.

Using the Pythagoras theorem,

(hypotenuse)² = (height)² + (base)²

=> (5x)² = (4x)² + (base)²

=> 25x² = 16x² + (base)²

=> (base)² = 25x² - 16x²

=> √(base)² = √(9x²)

=> base = 3x

.

Now,

$\bf cos = \frac{base}{hypotenuse } \\ \\ \bf = \frac{3x}{5x} \\ \\ \bf = \frac{3}{5} \\ \\ \bf so \\ \bf {cos}^{2} = { \bigg( \frac{3}{5} \bigg) }^{2} \\ \\ \bf = \frac{9}{25}$

Hence,

sin² = 16/25 & cos² = 9/25.

.

Hope it helps.

#BeBrainly :-)