`cos(pi - theta) + sin(pi/2 + theta) = 0` Prove the identity.

Monday, November 3, 2008

$\displaystyle{\cos{{\left(\pi-\theta\right)}}}+{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}={0}$ Prove the identity.

You need to use the formula $\displaystyle{\cos{{\left({a}-{b}\right)}}}={\cos{{a}}}\cdot{\cos{{b}}}+{\sin{{a}}}\cdot{\sin{{b}}}$ and $\displaystyle{\sin{{\left({a}+{b}\right)}}}={\sin{{a}}}\cdot{\cos{{b}}}+{\sin{{b}}}\cdot{\cos{{a}}},$ such that:

$\displaystyle{\cos{{\left(\pi-\theta\right)}}}={\cos{\pi}}\cdot{\cos{\theta}}+{\sin{\pi}}\cdot{\sin{\theta}}$

Since $\displaystyle{\cos{\pi}}=-{1}$ and $\displaystyle{\sin{\pi}}={0}$ , yields:

$\displaystyle{\cos{{\left(\pi-\theta\right)}}}=-{\cos{\theta}}$

$\displaystyle{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}={\sin{{\left(\frac{\pi}{{2}}\right)}}}\cdot{\cos{\theta}}+{\sin{\theta}}\cdot{\cos{{\left(\frac{\pi}{{2}}\right)}}}$

Since $\displaystyle{\sin{{\left(\frac{\pi}{{2}}\right)}}}={1}$ and $\displaystyle{\cos{{\left(\frac{\pi}{{2}}\right)}}}={0},$ yields:

$\displaystyle{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}={\cos{\theta}}$

Replacing $\displaystyle-{\cos{\theta}}$ for $\displaystyle{\cos{{\left(\pi-\theta\right)}}}$ and $\displaystyle{\cos{\theta}}$ for in $\displaystyle{\left(\frac{\pi}{{2}}+\theta\right)}$ yields:

$\displaystyle{\cos{{\left(\pi-\theta\right)}}}+{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}=-{\cos{\theta}}+{\cos{\theta}}={0}$

Hence, checking if the identity is valid yields that $\displaystyle{\cos{{\left(\pi-\theta\right)}}}+{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}={0}$ holds.

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Monday, November 3, 2008

$\displaystyle{\cos{{\left(\pi-\theta\right)}}}+{\sin{{\left(\frac{\pi}{{2}}+\theta\right)}}}={0}$ Prove the identity.

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