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cos300-sin(90-B) --------------- cos(-B)+sin330 how would this be solved using identities
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please anyone
use these : sin(90-B) = cosB cos(-B) = cosB sin(-B) = -sinB cos(x + 360) = cos(x) sin(x + 360) = sin(x)
i have but i get stuck here \[[(0.5)-cosB]/[cosb-1/2]\]
wonderful! one more step you're done !!
ONE MOAR
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ROW ROOW ROOWW YOUR BOAT
GENTLY DOWWNN THE STREAM
lol dan
moral support
\( [(0.5)-cosB]/[cosb-1/2] \) \(-1[cosB - 0.5]/[cosb-1/2] \)
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is that factorisation
no, i just pulled -1 out
(a-b) = -(b-a)
ooh then you can cross cancel
exactly !
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ooh thank you so much
np :)
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