Question

If angle=45 degree, then the value of cos^2(45)-sin^2(45)=?

Answer 1

Answer:

Ans:

cos^2(45) = 1/2

sin^2(45) =1/2

cos^(45) - sin^(45) = 1/2 - 1/2

= 0

Answer = 0

Answer 2

Topic:-

Trigonometry

Question:-

$\theta =45° , then find the Value of cos²\theta-sin²\theta =?$

Solution:-

$As we know that cos²\theta-sin²\theta = cos2\theta$

$so, substitute the \theta value in the formula$

$\theta= 45°$

$So, cos2\theta = cos2×45$

$cos2\theta=cos90$

$We know that cos90 = 0$

Answer:-

$cos2\theta=0$

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Multiples:-

$sin2\theta= 2sin\theta cos\theta$

$sin2\theta=\dfrac{2tan\theta}{1+tan²\theta}$

$cos2\theta= cos²\theta-sin²\theta$

$cos2\theta= 1-2sin²\theta$

$cos2\theta= 2cos²\theta-1$

$cos2\theta= \dfrac{1-tan²\theta}{1+tan²\theta}$

$tan2\theta= \dfrac{2tan\theta}{1-tan²\theta}$

$cot2\theta= \dfrac{cot²\theta-1}{2cot\theta}$