तीन कैन लिस्ट फर्स्ट थिंग यू डू टू हेल्प एट होम
Answers
Answer:
Consider unit circle with centre O at the origin let A be the point (1 , 0). Let P , Q and R be the points on the circle such that arc AP = x , arc PQ = y and arc AR = - y.
Then Arc AQ = arc AP + arc PQ = x + y.
Therefore, the co-ordinates of point P , Q and R are
\begin{gathered} \sf (cos \: x ,sin \: x),(cos(x + y) ,sin(x + y)) \\ \bf and \\ \sf (cos( - y),sin ( - y)) \: respectively\end{gathered}
(cosx,sinx),(cos(x+y),sin(x+y))
and
(cos(−y),sin(−y))respectively
We have ,
arc PQ = arc RA
⟼ arc PQ + arc AP = arc RA + arc AP
⟼ arc AQ = arc RP
⟼ Length of chord AQ = Length of chord RP
(∵ In a circle , equal arcs cut off equal chords)
⟼ AQ = RP = \sf AQ^2=RP^2AQ
2
=RP
2
\begin{gathered}\longrightarrow\sf( \cos(x + y) - 1) {}^{2} + (\sin(x + y - 0 {)}^{2} ) \\ \sf = ( \cos \:x - \cos( - y)) {}^{2} + (\sin \: x - \sin( - y) ) {}^{2} \end{gathered}
⟶(cos(x+y)−1)
2
+(sin(x+y−0)
2
)
=(cosx−cos(−y))
2
+(sinx−sin(−y))
2
\begin{gathered}\longrightarrow \sf { \cos }^{2} (x + y) + 1 - 2 \cos(x + y) + \sin {}^{2} (x + y) \\ \sf = {( \cos \: x - \cos \: y)}^{2} + ( { \sin \: x + \sin \: y)}^{2} \end{gathered}
⟶cos
2
(x+y)+1−2cos(x+y)+sin
2
(x+y)
=(cosx−cosy)
2
+(sinx+siny)
2
\bf sin ( - y) = - sin \: y \: \: and \: \: cos( - y) = cos \: ysin(−y)=−sinyandcos(−y)=cosy
\begin{gathered}\longrightarrow\sf \{\cos {}^{2} (x + y) + \sin {}^{2} (x + y) \} \\ \sf+ 1 - 2 \cos(x + y) \\ \sf = { \cos}^{2} x + \cos {}^{2} y - 2 \cos \: x \: \cos \: y \\ \sf + \: { \sin}^{2} x + { \sin}^{2} y + 2 \sin \: x \: \sin \: y\end{gathered}
⟶{cos
2
(x+y)+sin
2
(x+y)}
+1−2cos(x+y)
=cos
2
x+cos
2
y−2cosxcosy
+sin
2
x+sin
2
y+2sinxsiny
\begin{gathered} \longrightarrow\sf1 + 1 - 2 \cos(x + y) \\ \sf = {( \cos }^{2} x + { \sin}^{2} x) + {( \cos }^{2} y + { \sin}^{2} y) \\ \sf - 2 \cos x \cos y + 2 \sin x \sin y \end{gathered}
⟶1+1−2cos(x+y)
=(cos
2
x+sin
2
x)+(cos
2
y+sin
2
y)
−2cosxcosy+2sinxsiny
\begin{gathered} \sf\longrightarrow2 - 2 \cos(x + y) = \\ \sf1 + 1 - 2 \cos x \cos y + 2 \sin x \sin y \\ \\ \sf⟼ - 2 \cos(x + y) = - 2 \cos x \cos y + 2 \sin x \sin y \\ \\ \purple{ \underline {\boxed{{\bf\longrightarrow\cos(x + y) = \cos x \cos y - \sin x \sin y } }}}\end{gathered}
⟶2−2cos(x+y)=
1+1−2cosxcosy+2sinxsiny
⟼−2cos(x+y)=−2cosxcosy+2sinxsiny
⟶cos(x+y)=cosxcosy−sinxsiny
Hence Proved !
___________________________
Proof of cos 2x = cos²x - sin²x :
\begin{gathered} \sf \cos(2x) = \cos(x + x) \\ \\ \bf \{using \: above \: formula \} \\ \\ = \sf \cos x \cos x - \sin x \sin x \\ \\ \large\purple{ \implies \underline {\boxed{{\bf cos 2x=cos {}^{2} x-sin {}^{2} x} }}}\end{gathered}
cos(2x)=cos(x+x)
{usingaboveformula}
=cosxcosx−sinxsinx
⟹
cos2x=cos
2
x−sin
2
x
Hence Proved !