Question

If sin A = 1/2 then find rhe value of 1 - cos ^2 A

Answer 1

Answer:

1 - cos^2A = sin^2A

= (1/2)^2

= 1/4

Hence the answer is 1/4.....

Answer 2

GivEn:-

• SinA = $\sf \dfrac{1}{2}$

To find:-

• 1 - cos²A

SoluTion:-

GivEn that,

★ SinA = $\sf \dfrac{1}{2} = \dfrac{P}{H}$

Therefore,

• Perpendicular(P) = 1
• Hypotenuse(H) = 2

☆ DIAGRAM:

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Now,

☯ Using Pythagoras theorem -

★ H² = P² + B²

$:\implies$ 2² = 1² + B²

$:\implies$ 4 = 1 + B²

$:\implies$ 4 - 1 = B²

$:\implies$ 3 = B²

☯ Taking sqrt both side -

$:\implies\sf \sqrt{3} = \sqrt{B^2}$

$:\implies\sf B = \sqrt{3}$

Therefore, Base of right angled triangle is $\bf \sqrt{3}$.

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Now we have to find value of cosA,

$:\implies$ cosA = $\sf \dfrac{B}{H}$

Therefore, cosA = $\sf \dfrac{ \sqrt{3}}{2}$

☯ Now, Put the value of cosA in -

$:\implies$ 1 - cos²A

$:\implies\sf 1 - \bigg( \dfrac{ \sqrt{3}}{2} \bigg)^2$

$:\implies\sf 1 - \dfrac{3}{4}$

$:\implies\sf \dfrac{4 - 3}{4}$

$:\implies\sf \dfrac{1}{4}$

$\dag$ Hence, value of 1 - cos²A is $\sf \dfrac{1}{2}$

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Similarly,

We can also find the value of 1 - cos²A -

As we know that,

1 - cos² = sin²A $\qquad\lgroup\bigg\bf sin^2A + cos^2A = 1 \bigg\rgroup$

We have the value of, sinA = $\sf \dfrac{1}{2}$

Put value of sinA in -

$:\implies\sf 1 - cos^2A = sin^2$

$:\implies\sf 1 - cos^2A = \bigg( \dfrac{1}{2} \bigg)$

$:\implies\sf 1 - cos^2A = \dfrac{1}{4}$

$\dag$ Hence, We get the value of 1 - cos²A direct using identity.

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[tex]\boxed{ \begin{minipage}{7 cm} Fundamental Trigonometric Identities \\ \\ $\sin^2\theta + \cos^2\theta=1 \\ \\ 1+\tan^2\theta = \sec^2\theta \\ \\ 1+\cot^2\theta = \text{cosec}^2 \, \theta$ \end{minipage} }[/tex]

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