Question

sec(x)+tan(x)=2+squareroot{5} then find sin(x)+cos(x)

Answer 1

sec $x + \tan x = 2 + \sqrt{5}$. Then find $\sin x + \cos x$

Solution.

Let's solve the equation for $x$:

$$\sec x + \tan x = 2 + \sqrt{5}$$

Rewrite it in another form:

$$\frac{1}{\cos x} + \frac{\sin x}{\cos x} = 2 + \sqrt{5}$$

Use the Weierstrass substitution:

$$\sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \quad \text{and} \quad \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}$$

Substitute $t = \tan \frac{x}{2}$. Then

$$\sin x = \frac{2t}{1 + t^2} \quad \text{and} \quad \cos x = \frac{1 - t^2}{1 + t^2}$$

Then our equation has the form:

$$\frac{1 + t^2}{1 - t^2} + \frac{2t}{1 - t^2} = 2 + \sqrt{5}$$ $$\frac{1 + t^2 + 2t}{1 - t^2} = 2 + \sqrt{5}$$ $$\frac{(1 + t)^2}{(1 - t)(1 + t)} = 2 + \sqrt{5}$$ $$\frac{1 + t}{1 - t} = 2 + \sqrt{5}, \quad t \neq -1$$

Solve the equation for $t$:

$$\frac{1 + t}{1 - t} - 2 - \sqrt{5} = 0$$ $$\frac{1 + t - 2 - \sqrt{5} + 2t + \sqrt{5}t}{1 - t} = 0$$

Multiply by $1 - t$:

$$-1 - \sqrt{5} + 3t + \sqrt{5}t = 0$$ $$(3 + \sqrt{5})t = 1 + \sqrt{5}$$ $$t = \frac{1 + \sqrt{5}}{3 + \sqrt{5}}$$

Substitute back for $t = \tan \frac{x}{2}$:

$$\tan \frac {x}{2} = \frac {1 + \sqrt {5}}{3 + \sqrt {5}}$$

Take the inverse tangent of both sides:

$$\frac {x}{2} = \arctan \left(\frac {1 + \sqrt {5}}{3 + \sqrt {5}}\right) + \pi k, \quad k \in \mathbb {Z}$$

Then

$$x = 2 \arctan \left(\frac {1 + \sqrt {5}}{3 + \sqrt {5}}\right) + 2 \pi k, \quad k \in \mathbb {Z}$$

So find $\sin x + \cos x$:

$$\begin{array}{l} \sin x + \cos x = \sin \left(2 \arctan \left(\frac {1 + \sqrt {5}}{3 + \sqrt {5}}\right) + 2 \pi k\right) + \cos \left(2 \arctan \left(\frac {1 + \sqrt {5}}{3 + \sqrt {5}}\right) + 2 \pi k\right) = \\ = \sin \left(2 \arctan \left(\frac {1 + \sqrt {5}}{3 + \sqrt {5}}\right)\right) + \cos \left(2 \arctan \left(\frac {1 + \sqrt {5}}{3 + \sqrt {5}}\right)\right) \end{array}$$

Let's calculate $\sin x$:

$$\frac {1 + \sqrt {5}}{3 + \sqrt {5}} = \frac {1 + \sqrt {5}}{3 + \sqrt {5}} \cdot \frac {3 - \sqrt {5}}{3 - \sqrt {5}} = \frac {3 + 2 \sqrt {5} - 5}{4} = \frac {\sqrt {5} - 1}{2}$$

Use this formula $\sin 2x = 2\sin x\cos x$:

$$\sin \left(2 \arctan \left(\frac {\sqrt {5} - 1}{2}\right)\right) = 2 \sin \left[ \arctan \left(\frac {\sqrt {5} - 1}{2}\right) \right] \cos \left[ \arctan \left(\frac {\sqrt {5} - 1}{2}\right) \right]$$

Then use compositions of trig and inverse trig functions:

$$\sin (\arctan x) = \frac {x}{\sqrt {1 + x ^ {2}}} \quad \text{and} \quad \cos (\arctan x) = \frac {1}{\sqrt {1 + x ^ {2}}}$$

We have

$$\begin{array}{l} 2 \sin \left[ \arctan \left(\frac {\sqrt {5} - 1}{2}\right) \right] \cos \left[ \arctan \left(\frac {\sqrt {5} - 1}{2}\right) \right] = 2 \cdot \frac {\left(\frac {\sqrt {5} - 1}{2}\right)}{\sqrt {1 + \left(\frac {3 - \sqrt {5}}{2}\right)}} \cdot \frac {1}{\sqrt {1 + \left(\frac {3 - \sqrt {5}}{2}\right)}} = \\ = 2 \frac {\left(\frac {\sqrt {5} - 1}{2}\right)}{1 + \left(\frac {3 - \sqrt {5}}{2}\right)} = 2 \cdot \frac {\sqrt {5} - 1}{5 - \sqrt {5}} = 2 \cdot \frac {1}{\sqrt {5}} = \frac {2}{\sqrt {5}} \end{array}$$

Similarly

Use the formula $\cos 2x = 1 - 2\sin^2 x$

$$\cos \left[ 2 \arctan \left(\frac {\sqrt {5} - 1}{2}\right) \right] = 2 \cos^ {2} \left[ \arctan \left(\frac {\sqrt {5} - 1}{2}\right) \right] - 1 = \frac {2}{1 + \left(\frac {3 - \sqrt {5}}{2}\right)} - 1 = \frac {4}{2 + 3 - \sqrt {5}} - 1 = \frac {4 - 5 + \sqrt {5}}{5 - \sqrt {5}} = \frac {1}{\sqrt {5}}$$

So

$$\sin x + \cos x = \frac {2}{\sqrt {5}} + \frac {1}{\sqrt {5}} = \frac {3}{\sqrt {5}}$$

Answer:

$$\sin x + \cos x = \frac {3}{\sqrt {5}}$$