Question

Express \[{\text{sin}}{67^0}{\text{ }} + {\text{ cos}}{75^0}\] in terms of trigonometric ratios of angles
between \[{0^0}\] and \[{45^0}\].

Answer 1

Hint:- Use \[{\text{6}}{{\text{7}}^0} = {90^0} - {23^0}\]and \[{75^0} = {90^0} - {15^0}\].

We are given with the equation,
\[ \Rightarrow {\text{sin}}{67^0}{\text{ }} + {\text{ cos}}{75^0}\] ---->(1)
And, had to convert the angles of the given equation between \[{0^0}\] and \[{45^0}\].
We know that,
According to the trigonometric identities,
\[ \Rightarrow {\text{sin}}\left( {{{90}^0} - \theta } \right) = \cos \theta \] ---->(2)
\[ \Rightarrow \]And, \[{\text{cos}}\left( {{{90}^0} - \theta } \right) = \sin \theta \] ----->(3)
So, above equation 1 can be written as,
\[ \Rightarrow \sin {\left( {90 - 23} \right)^0} + \cos {\left( {90 - 15} \right)^0}\] ---->(4)
So, using equation 2 and 3. We can write equation 4 as,
\[ \Rightarrow {\text{cos}}{23^0} + {\text{sin1}}{5^0}\]
Hence, the given equation is expressed in trigonometric ratios
of angles lying between \[{0^0}\] and \[{45^0}\].
\[ \Rightarrow \]So, \[{\text{sin}}{67^0}{\text{ }} + {\text{ cos}}{75^0} = {\text{cos}}{23^0} + {\text{sin1}}{5^0}\]
Note:- Whenever we came up with this type of problem then remember that,
\[{\text{sin}}\left( {{{90}^0} - \theta } \right) = \cos \theta \] and \[{\text{cos}}\left( {{{90}^0} - \theta } \right) = \sin \theta \]. And value of\[{\text{ sin}}\theta ,{\text{ cos}}\theta ,{\text{ tan}}\theta ,\]
\[{\text{cot}}\theta ,{\text{ cosec}}\theta \] and \[{\text{sec}}\theta \] is always positive if \[\theta \in \left[ {0,{{90}^0}} \right].\]