How do you use the sum or difference identities to find the exact value of #sec1275^circ#?

Question

2869 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-09T08:47:03

#sec(1275)=sqrt2-sqrt6#

#sec(1275)=1/cos(1275)#

now cosine is periodic with period #360^0#

#=1/cos915=1/cos555=1/cos195#

now #195^0# is in the third quadrant, where cosine is negative, so we have

#=-1/cos15---(1)#

we can now evaluate #cos15#

#cos15=cos(45-cos30)#

#=cos45cos30+sin45sin30#

using the values for the standard angles

#=sqrt2/2sqrt3/2+sqrt2/2 1/2#

#=(sqrt6+sqrt2)/4#

substituting into #(1)#

#1/sec(1275)=-1/cos15=-4/(sqrt6+sqrt2)#

rationalising

#=-4(sqrt6-sqrt2)/4=-(sqrt6-sqrt2)=sqrt2-sqrt6#