If the value of sin A is 45- then what will be the value of sin 2 A-cos2 A -

Given : sin A =45=4k5k (Let) Now, perpendicular = 4k and hypotenuse = 5k. So, base = √((5k)^2 (4k)^2 )= √(25k^2 16k^2) ⇒ base=√(9k^2)=3k Finally, cos A = ba ...

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Standard XII

Mathematics

Trigonometric Ratios of Half Angles

If the value ...

Question

If the value of sin A is $\frac{4}{5}$ , then what will be the value of ${sin}^{2} A + {cos}^{2} A$ ?

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Solution

Given : sin A $= \frac{4}{5} = \frac{4 k}{5 k}$ (Let)
Now, perpendicular = 4k and hypotenuse = 5k.
So, base = $\sqrt{(5 k)^{2} - (4 k)^{2}} = \sqrt{25 k^{2} - 16 k^{2}} \Rightarrow b a s e = \sqrt{9 k^{2}} = 3 k$
Finally, $cos A = \frac{b a s e}{h y p o t e n u s e} = \frac{3 k}{5 k} = \frac{3}{5}$
Thus, ${sin}^{2} A + {cos}^{2} A = {(\frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}$
$\Rightarrow \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1$

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If the value of sin A is 45, then what will be the value of sin2A+cos2A ?

Given : sin A =45=4k5k (Let) Now, perpendicular = 4k and hypotenuse = 5k. So, base = √(5k)2−(4k)2=√25k2−16k2⇒ base=√9k2=3k Finally, cos A=basehypotenuse=3k5k=35 Thus, sin2A+cos2A=(45)2+(35)2 ⇒ 1625+925=2525=1

If the value of sin A is $\frac{4}{5}$ , then what will be the value of ${sin}^{2} A + {cos}^{2} A$ ?