Question

PROVE THAT:

Sin²π/6 + Cos²π/3 - Tan²π/4 = -1/2

Answer 1

= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
= 35

an²(π/16) + tan²(7π/16) 
= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)

Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
= 35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)

Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3

= 35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)

Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
= 35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)

Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
= 35

an²(π/16) + tan²(7π/16) 
= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)

Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3

= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
= 35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)

Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
= 35

2 years agoApproved

Answer 2

sin π/6 = sin 30° = 1/2

cos π/3 = cos 60° = 1/2

tan π/4 = tan 45° = 1

thus
sin²π/6 +cos²π/3 -tan²π/4

= (1/2)²+(1/2)²-(1)²

= 1/4 +1/4 -1

= 1/2 -1

= -1/2

hence proved