If #8cosx + 15sinx = 15# and #cosx!=0# then which of the following option is correct for #8sinx-15cosx#?

#a)8#
#b)-8#

2 Answers
Ratnaker Mehta
Jan 13, 2018

#b) -8.#

Explanation:

Given that, #8cosx+15sinx=15#.

#rArr 15sinx=15-8cosx#.

#rArr 225sin^2x=225-240cosx+64cos^2x....[because," squaring]"#.

#rArr 225(1-cos^2x)=225-240cosx+64cos^2x#.

#rArr225-225cos^2x=225-240cosx+64cos^2x#.

#rArr289cos^2x-240cosx=0#

#rArr cosx(289cosx-240)=0#.

#rArr cosx=0, or, cosx=240/289#.

But, we are given that, #cosx ne 0, :. cosx=240/289#.

Utilising this value of #cosx# in #15sinx=15-8cosx#, we get,

#sinx=1/15(15-8cosx)=1-8/15*240/289=1-128/289=161/289#

#:."The Reqd. Val.="8sinx-15cosx#,

#=8*161/289-15*240/289#,

#=(1288-3600)/289#,

#=-2312/289#.

#rArr 8sinx-15cosx=-8#.

Therefore, #b) -8# is the right option.

P dilip_k
Jan 13, 2018

Given

#8cosx+15sinx=15......(1)#

Now

#(8sinx-15cosx)^2+(8cosx+15sinx)^2=8^2(sin^2x+cos^2x)+15^2(cos^2x+sin^2x)-2*8*15sinxcosx+2*8*15sinxcosx#

#=>(8sinx-15cosx)^2+15^2=8^2+15^2#

#=>(8sinx-15cosx)^2=8^2#

#=>8sinx-15cosx=pm8#

So

#8sinx-15cosx=8........(2)#

and

#8sinx-15cosx=-8........(3)#

Now we are to find out which one between (2) and (3) is true.

Adding (1) and (2) we get

#23sinx-7cosx=23# This relation is possible only if #cosx=0# but given condition is #cosx!=0#

So relation (2) is rejected and relation (3) is accepted.

This means option (1) is correct.

Related questions
Impact of this question
6016 views around the world
You can reuse this answer
Creative Commons License