Question

Prove that cos50° + cos70° = sin80°.

Answer 1

Hint: We know that cos (C) + cos (D) = 2$\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$. Hence we will use this formula to evaluate the given equation. After this we will get cos10 and cos60 in our equation.

Complete step by step answer:
Now we know that cos 60 = $\dfrac{1}{2}$ and cos (θ) = sin (90-θ).
Now we need to prove that cos50° + cos70° = sin80°.
Now let us consider left hand side which is cos50° + cos70°
Now we know that cos (C) + cos (D) = 2$\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$.
We will use this formula to evaluate cos (C) + cos (D) = 2$\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$.
Let us take C = 50° and D = 70°.
Hence cos(C) + cos(D) = cos(50°) + cos(70°)
Hence we get
cos(50°) + cos(70°) = $2\cos \left( \dfrac{50+70}{2} \right)\cos \left( \dfrac{50-70}{2} \right)$
Now 50 + 70 = 120 and 50 – 70 = -20, hence we get
cos(50°) + cos(70°) = $2\cos \left( \dfrac{120}{2} \right)\cos \left( \dfrac{-20}{2} \right)$
hence we get
cos(50°) + cos(70°) = 2 cos(60°)cos(-10°)
Now we know that cos(60°) = $\dfrac{1}{2}$ Hence we get
cos(50°) + cos(70°) = cos(-10°)
Now using and cos(-θ) = cos(θ) we get cos(-10°) = cos(10°)
Hence we have
cos(50°) + cos(70°) = cos(10°)
Now we also know that cos(θ) = sin(90 - θ). Hence using this in the above equation we can write cos(10°) as sin(90-10)
Hence we get
cos(50°) + cos(70°) = sin(90 – 10)
cos(50°) + cos(70°) = sin(80°)
Hence the given equation is proved.
Note:
Note that in the formula cos (C) + cos (D) = 2$\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ the value does not change if we take C as 70 and D as 50, this is because of the fact that $\cos (-\theta )=\cos (\theta )$