Question

Cos55+cos65+cos175=0

Answer 1

Question :-

Prove that

$\rm \: cos55\degree + cos65\degree + cos175\degree = 0$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: cos55\degree + cos65\degree + cos175\degree$

can be re-arranged as

$\rm \: = \: cos65\degree + cos55\degree + cos175\degree$

We know,

$\boxed{\sf{  \:\rm \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: \: }}$

So, on this result, we get

$\rm \: = \:2cos\bigg[\dfrac{65\degree + 55\degree }{2} \bigg]cos\bigg[\dfrac{65\degree - 55\degree }{2} \bigg] + cos175\degree$

$\rm \: = \:2cos\bigg[\dfrac{120\degree }{2} \bigg]cos\bigg[\dfrac{10\degree }{2} \bigg] + cos175\degree$

$\rm \: = \:2cos60\degree cos5\degree + cos175\degree$

$\rm \: = \:2 \times \frac{1}{2} \times cos5\degree + cos175\degree$

$\rm \: = \:cos5\degree + cos175\degree$

can be further rewritten as

$\rm \: = \:cos5\degree + cos(180\degree - 5\degree )$

We know,

$\boxed{\sf{  \:\rm \: cos(180\degree - x) \: = - \: cosx \: \: }}$

So, using this result, we get

$\rm \: = \:cos5\degree - cos5\degree$

$\rm \: = \:0$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: cos55\degree + cos65\degree + cos175\degree = 0 \: \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\boxed{\sf{  \:\rm \: sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: \: }}$

$\boxed{\sf{  \:\rm \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: \: }}$

$\boxed{\sf{  \:\rm \: cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg] \: \: }}$

$\boxed{\sf{  \:\rm \: 2sinxcosy = sin(x + y) + sin(x - y) \: \: }}$

$\boxed{\sf{  \:\rm \: 2cosxcosy = cos(x + y) + cos(x - y) \: \: }}$

$\boxed{\sf{  \:\rm \: 2sinxsiny = cos(x - y) + cos(x + y) \: \: }}$

Answer 2

Answer:

$</p><p></p><p>\begin{gathered}\rm \: cos55\degree + cos65\degree + cos175\degree = 0 \\ \end{gathered}cos55°+cos65°+cos175°=0</p><p></p><p>$

$\large\underline{\sf{Solution-}}</p><p></p><p>$

$\rm \: Consider LHS</p><p></p><p>$

$\begin{gathered}\rm \: cos55\degree + cos65\degree + cos175\degree \\ \end{gathered}cos55°+cos65°+cos175° \\ </p><p></p><p>$

$\rm \: can \: be \: re-arranged \: as</p><p></p><p>$

$\begin{gathered}\rm \: = \: cos65\degree + cos55\degree + cos175\degree \\ \end{gathered} \tt \: =cos65°+cos55°+cos175°</p><p></p><p>$

$\rm \: We \: know,</p><p></p><p>\begin{gathered}\boxed{\sf{  \:\rm \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: \: }} \\ \end{gathered} cosx+cosy=2cos[2x+y]cos[2x−y]</p><p></p><p>$

$\rm \: So, on \: this \: result, \: we \: get</p><p></p><p>\rm \: = \:2cos\bigg[\dfrac{65\degree + 55\degree }{2} \bigg]cos\bigg[\dfrac{65\degree - 55\degree }{2} \bigg] + cos175\degree=2cos[265°+55°]cos[265°−55°]+cos175°</p><p></p><p>\begin{gathered}\rm \: = \:2cos\bigg[\dfrac{120\degree }{2} \bigg]cos\bigg[\dfrac{10\degree }{2} \bigg] + cos175\degree \\ \end{gathered}=2cos[2120°]cos[210°]+cos175°</p><p></p><p>\begin{gathered}\rm \: = \:2cos60\degree cos5\degree + cos175\degree \\ \end{gathered}=2cos60°cos5°+cos175°</p><p></p><p>\begin{gathered}\rm \: = \:2 \times \frac{1}{2} \times cos5\degree + cos175\degree \\ \end{gathered}=2×21×cos5°+cos175°</p><p></p><p>\begin{gathered}\rm \: = \:cos5\degree + cos175\degree \\ \end{gathered}=cos5°+cos175°</p><p></p><p>can \: be \: further \: rewritten \: as</p><p></p><p>\begin{gathered}\rm \: = \:cos5\degree + cos(180\degree - 5\degree ) \\ \end{gathered}=cos5°+cos(180°−5°)</p><p></p><p>We know,</p><p></p><p>\begin{gathered}\boxed{\sf{  \:\rm \: cos(180\degree - x) \: = - \: cosx \: \: }} \\ \end{gathered} cos(180°−x)=−cosx</p><p></p><p>So, using this result, we get</p><p></p><p>\begin{gathered}\rm \: = \:cos5\degree - cos5\degree \\ \end{gathered}=cos5°−cos5°</p><p></p><p>\begin{gathered}\rm \: = \:0 \\ \end{gathered}=0</p><p></p><p>Hence,</p><p></p><p>\begin{gathered}\rm\implies \:\boxed{\sf{  \:\rm \: cos55\degree + cos65\degree + cos175\degree = 0 \: \: }} \\ \end{gathered}⟹ cos55°+cos65°+cos175°=0</p><p></p><p></p><p>$