Prove the trigonometric identity below. Show all steps algebraically
(š„sinĪøāš¦cosĪø)2 +(š„cosĪø+š¦sinĪø)2 =š„2 +š¦2
4 answers
the 2's are exponents
Someone else posted that earlier:
www.jiskha.com/questions/1866365/solve-the-following-equation-and-state-the-general-solution-for-all-values-of-x-in-exact
www.jiskha.com/questions/1866365/solve-the-following-equation-and-state-the-general-solution-for-all-values-of-x-in-exact
Not the same question, so here it goes
prove: (š„sinĪøāš¦cosĪø)^2 +(š„cosĪø+š¦sinĪø)^2 =š„^2 +š¦^2
Let x = sinA and let y = cosA
Left Side = (š„sinĪøāš¦cosĪø)^2 +(š„cosĪø+š¦sinĪø)^2
= (sinAsinĪø - cosAcosĪø) + (sinAcosĪø + cosAsinĪø)^2
= ( cos(A+Īø))^2 + (sin(A+Īø))^2
= 1
RS = (sinA)^2 + (cosA)^2
= 1
= LS
prove: (š„sinĪøāš¦cosĪø)^2 +(š„cosĪø+š¦sinĪø)^2 =š„^2 +š¦^2
Let x = sinA and let y = cosA
Left Side = (š„sinĪøāš¦cosĪø)^2 +(š„cosĪø+š¦sinĪø)^2
= (sinAsinĪø - cosAcosĪø) + (sinAcosĪø + cosAsinĪø)^2
= ( cos(A+Īø))^2 + (sin(A+Īø))^2
= 1
RS = (sinA)^2 + (cosA)^2
= 1
= LS
clever solution, but if |x| or |y| >1 then they cannot be sin and cos of A. In that case, some more algebra will be needed, but it still works out to x^2 + y^2.