Question

sin 78degres-sin18degrees+cos132degrees=0
​

Answer 1

Answer:

$sin78° - sin18° + cos132°</p><p></p><p>\begin{gathered}2cos( \frac{78 \degree + 18 \degree}{2} ).sin( \frac{78 \degree- 18 \degree}{2} ) + cos132 \degree \\ \\ = > 2cos( \frac{96}{2} ).sin( \frac{70 \degree - 18 \degree}{2} ) + cos132 \degree \\ \\ = > 2cos48 \degree.sin30 \degree + cos132 \degree \\ \\ = > 2cos48 \degree.( \frac{1}{2} ) + cos132 \degree \\ \\ = > cos48 \degree + cos132 \degree \\ \\ = > 2cos( \frac{48 \degree + 132 \degree}{2} ).cos( \frac{48 \degree - 132 \degree}{2} ) \\ \\ = > 2.cos( \frac{180 \degree}{2} ).cos( \frac{ - 84}{2} ) \\ \\ = > 2.cos90 \degree.cos( - 42 \degree) \\ \\ = > 2.(0).cos( - 42) \\ \\ = > 0.cos ( - 42) \\ \\ = > 0 \: \: \: \: \: \: \: \: \: [R.H.S.]\end{gathered}2cos(278°+18°).sin(278°−18°)+cos132°=>2cos(296).sin(270°−18°)+cos132°=>2cos48°.sin30°+cos132°=>2cos48°.(21)+cos132°=>cos48°+cos132°=>2cos(248°+132°).cos(248°−132°)=>2.cos(2180°).cos(2−84)=>2.cos90°.cos(−42°)=>2.(0).cos(−42)=>0.cos(−42)=>0[R.H.S.]</p><p></p><p></p><p>$

hope you like it

plz mark it as the brainliest