Question

$\text{If}\sin \left({\theta }_1+{\theta }_2\right)=1\text{and}\cos \left({\theta }_1-{\theta }_2\right)=1,0°<{\theta }_1+{\theta }_2\le 90°,{\theta }_1\ge {\theta }_2$then find ${\theta }_1\text{and}{\theta }_2$.

Answer 1

Given data: $\sin \left({\theta }_1+{\theta }_2\right)=1$ and $\cos \left({\theta }_1-{\theta }_2\right)=1$

Finding the value of ${\theta }_1+{\theta }_2$:

Since,

$$\begin{gathered} \sin ({\theta }_1+{\theta }_2)=1 \\ \Rightarrow \sin ({\theta }_1+{\theta }_2)=\sin \frac{\pi }{2} \\ \Rightarrow {\theta }_1+{\theta }_2=\frac{\pi }{2} \\ \therefore {\theta }_1+{\theta }_2=\frac{\pi }{2}.......\left(1\right) \end{gathered}$$

Finding the value of ${\theta }_1-{\theta }_2$:

Since,

$$\begin{gathered} \cos ({\theta }_1-{\theta }_2)=1 \\ \Rightarrow \cos ({\theta }_1-{\theta }_2)=\cos 0^o \\ \Rightarrow {\theta }_1-{\theta }_2=0^o \\ \therefore {\theta }_1-{\theta }_2=0^o.........\left(2\right) \end{gathered}$$

Adding equation (1) and (2) we get,

$$\begin{gathered} ({\theta }_1+{\theta }_2)+({\theta }_1-{\theta }_2)=\frac{\pi }{2}+0^o \\ \Rightarrow {\theta }_1+{\theta }_2+{\theta }_1-{\theta }_2=\frac{\pi }{2} \\ \Rightarrow 2{\theta }_1=\frac{\pi }{2} \\ \Rightarrow {\theta }_1=\frac{\pi }{4} \\ \therefore {\theta }_1=\frac{\pi }{4} \end{gathered}$$

And Substituting the value of ${\theta }_1$ in equation (1)

$$\begin{gathered} {\theta }_1+{\theta }_2=\frac{\pi }{2} \\ \Rightarrow {\theta }_2=\frac{\pi }{2}-{\theta }_1 \\ \Rightarrow {\theta }_2=\frac{\pi }{2}-\frac{\pi }{4} \\ \Rightarrow {\theta }_2=\frac{\pi }{4} \\ \therefore {\theta }_2=\frac{\pi }{4} \end{gathered}$$

Hence, the values of ${\theta }_1$and ${\theta }_2$is $\frac{\mathrm{\pi }}{4}$and $\frac{\mathrm{\pi }}{4}$respectively.