Question

tan⁡θ= (sin⁡α-cos⁡α)/(sin⁡α+cos⁡α ) ,then sin⁡α+cos⁡α = ?

Answer 1

Answer on Question #44432 – Math – Trigonometry

$$\tan(\theta) = \frac{\sin(\alpha) - \cos(\alpha)}{\sin(\alpha) + \cos(\alpha)} \quad \text{Then} \quad \sin(\alpha) + \cos(\alpha) = ?$$

Solution

$$\tan(\theta) = \frac{\tan(\alpha) - 1}{\tan(\alpha) + 1} = 1 - \frac{2}{\tan(\alpha) + 1}.$$

Rewrite

$$\frac{2}{\tan(\alpha) + 1} = 1 - \tan(\theta),$$ $$\tan(\alpha) + 1 = \frac{2}{1 - \tan(\theta)}$$ $$\tan(\alpha) = \frac{2}{1 - \tan(\theta)} - 1$$ $$\tan(\alpha) = \frac{2 - 1 + \tan(\theta)}{1 - \tan(\theta)}$$ $$\tan(\alpha) = \frac{1 + \tan(\theta)}{1 - \tan(\theta)}$$ $$\tan^2(\alpha) = \left(\frac{1 + \tan(\theta)}{1 - \tan(\theta)}\right)^2$$ $$\tan^2(\alpha) + 1 = \left(\frac{1 + \tan(\theta)}{1 - \tan(\theta)}\right)^2 + 1 = \frac{\left(1 + \tan(\theta)\right)^2 + \left(1 - \tan(\theta)\right)^2}{\left(1 - \tan(\theta)\right)^2} = \frac{2\left(1 + \tan^2(\theta)\right)}{\left(1 - \tan(\theta)\right)^2}$$

It is known that $\tan^2(\alpha) + 1 = \frac{1}{\cos^2(\alpha)}$, hence

$\cos^2(\alpha) = \frac{1}{\tan^2(\alpha) + 1}$, substitute expression (1) and obtain

$$\cos^2(\alpha) = \frac{\left(1 - \tan(\theta)\right)^2}{2\left(1 + \tan^2(\theta)\right)}.$$

It is known

$$\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{\left(1 - \tan(\theta)\right)^2}{2\left(1 + \tan^2(\theta)\right)} = \frac{2\left(1 + \tan^2(\theta)\right) - \left(1 - \tan(\theta)\right)^2}{2\left(1 + \tan^2(\theta)\right)}$$ $$\sin^2(\alpha) = \frac{1 + \tan^2(\theta) + 2\tan(\theta)}{2\left(1 + \tan^2(\theta)\right)} = \frac{\left(1 + \tan(\theta)\right)^2}{2\left(1 + \tan^2(\theta)\right)}$$

If $\sin(\alpha) > 0$, $\cos(\alpha) > 0$, then

$$\sin(\alpha) + \cos(\alpha) = \sqrt{ \frac{ (1 + \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } + \sqrt{ \frac{ (1 - \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } = \frac{ |1 + \tan(\theta)| + |1 - \tan(\theta)| }{ \sqrt{ 2(1 + \tan^2(\theta)) } }$$

If $\sin(\alpha) > 0$, $\cos(\alpha) < 0$, then

$$\sin(\alpha) + \cos(\alpha) = \sqrt{ \frac{ (1 + \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } - \sqrt{ \frac{ (1 - \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } = \frac{ |1 + \tan(\theta)| - |1 - \tan(\theta)| }{ \sqrt{ 2(1 + \tan^2(\theta)) } }$$

If $\sin(\alpha) < 0$, $\cos(\alpha) > 0$, then

$$\sin(\alpha) + \cos(\alpha) = - \sqrt{ \frac{ (1 + \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } + \sqrt{ \frac{ (1 - \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } = \frac{ - |1 + \tan(\theta)| + |1 - \tan(\theta)| }{ \sqrt{ 2(1 + \tan^2(\theta)) } }$$

If $\sin(\alpha) < 0$, $\cos(\alpha) < 0$, then

$$\sin(\alpha) + \cos(\alpha) = - \sqrt{ \frac{ (1 + \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } - \sqrt{ \frac{ (1 - \tan(\theta))^2 }{ 2(1 + \tan^2(\theta)) } } = \frac{ - |1 + \tan(\theta)| - |1 - \tan(\theta)| }{ \sqrt{ 2(1 + \tan^2(\theta)) } }$$

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