# Solving trigonometric equations

If we have $$\sin (A) +\cos (A) + \csc (A) + \sec (A) +\tan (A) +\cot (A)= 7$$ and $$\sin(2A) =a-b\sqrt{7}= 2\sin(A)\cos(A).$$ What values can $a$ and $b$ take?

Let $A=x$ and $\sin{x}+\cos{x}=t$.

Hence, $|t|\leq\sqrt2$, $\sin{x}\cos{x}=\frac{t^2-1}{2}$ and we need to solve $$t+\frac{t}{\frac{t^2-1}{2}}+\frac{1}{\frac{t^2-1}{2}}=7$$ or $$t+\frac{2}{t-1}=7$$ or $$t^2-8t+9=0$$ or $$(t-4)^2=7,$$ which gives $t=4+\sqrt7$, which is impossible, or $t=4-\sqrt7$,

which gives $\sin2x=t^2-1=(4-\sqrt7)^2-1=22-8\sqrt7$.

Id est, $a=22$ and $b=8$ if you mean that $a$ and $b$ are naturals.

Done!

• I still did'nt understand one thing that why is |t| less than equal to √2 ? Could you please give me a simple explanation Jul 26 '17 at 15:27
• @Albert Pinto because $\sin{x}+\cos{x}=\sqrt2\left(\sin{x}\cos45^{\circ}+\cos{x}\sin45^{\circ}\right)=\sqrt2\sin\left(x+45^{\circ}\right)$ and $-1\leq \sin\leq1$. Jul 26 '17 at 15:34
• @Albert Pinto Also there is the following explanation. Let $\sin{x}=a$ and $\cos{x}=b$. Hence, $a^2+b^2=1$ and we need to prove that $|a+b|\leq\sqrt2$ or $(a+b)^2\leq2$ or $(a+b)^2\leq2(a^2+b^2)$ or $(a-b)^2\geq0$, which is obvious. Jul 26 '17 at 15:46

HINT: $$\sin(2A)=22-8\sqrt{7}$$