Question

Prove the given trigonometric equation as :
${\text{Sin2}}{{\text{3}}^0} + {\text{Sin3}}{{\text{7}}^0} = {\text{Cos}}{{\text{7}}^0}.$

Answer 1

Hint: In this type of question, where sum of two trigonometric function is given and angles are non standard then in such case you should use sum to product identity of sine function which is given as: ${\text{SinC + SinD = 2Sin}}\left( {\dfrac{{{\text{C + D}}}}{2}} \right){\text{Cos}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right).$
Complete step-by-step solution -
This is a simple problem from chapter trigonometric identities.
In this question, we have to prove the given expression. So, the usual approach to dealing with this type of question is to first consider the expression in Left Hand Side(LHS) and solve it using different formulas in such a way that it becomes equal to the expression given in Right Hand Side(RHS).
i.e. ${\text{LHS = RHS}}$ .
So, accordingly we will first consider LHS.
In the LHS, it is given the sum of sine of two angles i.e. ${\text{Sin2}}{{\text{3}}^0} + {\text{Sin3}}{{\text{7}}^0}$

$\because$ The given angles are not standard angles. So, we have to use trigonometric identities to further solve this question.
According to sum to product identity of sine function:
${\text{SinC + SinD = 2Sin}}\left( {\dfrac{{{\text{C + D}}}}{2}} \right){\text{Cos}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right).$ (1)
Where C and D are angles in degree or radian.
Now from the question, we can easily understand the value of angle C=230 and D=370.
Putting these values in expression given in LHS, we get:
${\text{Sin2}}{{\text{3}}^0} + {\text{Sin3}}{{\text{7}}^0}$
Now using the above identity given in equation(1), we get:${\text{Sin2}}{{\text{3}}^0} + \operatorname{Sin} {37^0} = 2{\text{Sin}}\left( {\dfrac{{{{23}^0} + {{37}^0}}}{2}} \right){\text{Cos}}\left( {\dfrac{{{{23}^0} - {{37}^0}}}{2}} \right) = 2{\text{Sin(3}}{{\text{0}}^0}){\text{Cos( - }}{{\text{7}}^0})$
$\because$ We know that${\text{Cos( - }}\theta {\text{) = Cos(}}\theta {\text{)}}{\text{.}}$

$\therefore$ The above expression can be written as:
${\text{Sin2}}{{\text{3}}^0} + {\text{Sin3}}{{\text{7}}^0} = 2{\text{Sin3}}{{\text{0}}^0}{\text{Cos}}{{\text{7}}^0} = 2 \times \dfrac{1}{2} \times {\text{Cos}}{{\text{7}}^0} = {\text{Cos}}{{\text{7}}^0}\varphi$
Therefore, the value of ${\text{LHS = Cos}}{{\text{7}}^0}.$
But, it is given that ${\text{RHS = Cos}}{{\text{7}}^0}$ .
Therefore, ${\text{RHS = LHS}}$
Hence, it is proved that ${\text{Sin2}}{{\text{3}}^0} + {\text{Sin3}}{{\text{7}}^0} = {\text{Cos}}{{\text{7}}^0}.$
NOTE: In such type of question where standard angles are not given, you should try to use
trigonometry identities. In this question we are using sum to product identity of sine function.
So, you should remember this formula. If you have forgot the formula, you can easily get it by
putting ${\text{C = A + B}}$ and ${\text{D = A - B}}$ and then replace A by $\dfrac{{{\text{C + D}}}}{2}$ and B by $\dfrac{{{\text{C - D}}}}{2}$