Question

If cos(α+β) = 0, then sin(α -β) can be reduced to:
(a) cosβ
(b) 2cosβ
(c) sinα
(d) 2sinα​

Answer 1

Answer:

The correct answer is

$\cos(2 \beta )$

Step-by-step explanation:

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Answer 2

Answer:

• cos(α+β) = 0
• sin(α-β) = ?

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

$\displaystyle\sf :\implies cos(\alpha+\beta) = 0$

• So the value with cos θ as 0 is none other than cos 90°

$\\\displaystyle\sf :\implies cos(\alpha+\beta) = cos 90^{\circ}$

$\displaystyle\sf :\implies \alpha+\beta = 90$

$\displaystyle\sf :\implies \alpha = 90-\beta\:\:\: -eq(1)$

• So now in sin(α-β) we shall substitute the value of α and on simple calculation we can get our answer!!

$\displaystyle\sf \dashrightarrow sin(\alpha-\beta)$

$\displaystyle\sf \dashrightarrow sin(90-\beta-\beta)$

$\displaystyle{\scriptsize\qquad\bf{\because}\:\:\texttt{ From eq(1)}}$

$\displaystyle\sf \dashrightarrow sin(90-2\beta)$

$\displaystyle{\scriptsize\qquad\bf{\because}\:\:\tt{ sin(90-\theta) = cos\theta}}$

$\displaystyle\sf \dashrightarrow\underline{\boxed{\sf cos2\beta}}$

$\displaystyle\therefore\:\underline{\textsf{We can reduce it to \textbf{ cos2}}\bf\beta}$

$\begin{gathered}\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}\end{gathered}$