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# what is the value of sin(37)cos(53)????angles in degree

PASUPULETI GURU MAHESH
126 Points
5 years ago
sin(90-37))The value of sin(37)cos(53) = sin(37)sin(90-37) (since cos(53)= sin(37)sin(37) =
= sin2(37)        (since a.a=a2)
we know that sin (37)= 0.60  then sin2 (37)= 0.36

Therefore the value of the sin(37)cos(53)=sin2(37)=0.36

Hansraj Gyanendra Singh Rajawat
35 Points
5 years ago
I want to know how the value of sin 37 came???? this is my actual question how you calculate this value???????????
Ajay
209 Points
5 years ago
Here is the solution..................................................................................................................

$Let\quad us\quad first\quad find\quad value\quad of\quad sin15\quad as\quad this\quad will\quad be\quad used\quad later.\\ we\quad know\quad cos30\quad =\quad 1-2{ sin }^{ 2 }15\\ This\quad will\quad give\quad us\quad sin15\quad =\quad \frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } \\ Now\quad use\quad the\quad formula\quad to\quad find\quad value\quad of\quad cos53sin37\quad as\quad follows\\ cos\alpha sin\beta \quad =\quad 1/2\left[ sin(\alpha +\beta )\quad -sin(\alpha -\beta ) \right] \\ Hence\quad cos53sin37\quad =\quad 1/2\left[ sin\quad 90\quad -sin\quad 15 \right] \\ Put\quad values\quad of\quad sin\quad 90\quad =\quad 1\quad and\quad sin15=\quad \frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } \\ cos53sin37\quad =\quad 1/2\left( 1-\frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } \right)$